Scalar or Dot product of two vectors and its applications Questions in English

(C) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are mutually orthogonal,their dot products must be zero.
First,check $\bar{a} \cdot \bar{b} = (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0$.
Next,use $\bar{a} \cdot \bar{c} = 0$:
$(1)(p) + (-1)(1) + (2)(q) = p - 1 + 2q = 0 \implies p + 2q = 1$ ... $(i)$
Then,use $\bar{b} \cdot \bar{c} = 0$:
$(2)(p) + (4)(1) + (1)(q) = 2p + 4 + q = 0 \implies 2p + q = -4$ ... $(ii)$
From $(ii)$,$q = -4 - 2p$. Substituting this into $(i)$:
$p + 2(-4 - 2p) = 1$
$p - 8 - 4p = 1$
$-3p = 9 \implies p = -3$.
Substituting $p = -3$ into $q = -4 - 2p$:
$q = -4 - 2(-3) = -4 + 6 = 2$.
Thus,$(p, q) = (-3, 2)$.

(B) Let $\theta$ be the angle between $\overline{a}$ and $\overline{b}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.
Given that $(5 \overline{a} + 4 \overline{b})$ and $(\overline{a} - 2 \overline{b})$ are perpendicular,their dot product is zero.
$(5 \overline{a} + 4 \overline{b}) \cdot (\overline{a} - 2 \overline{b}) = 0$
$5(\overline{a} \cdot \overline{a}) - 10(\overline{a} \cdot \overline{b}) + 4(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$
$5|\overline{a}|^2 - 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$
Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,we have $5(1)^2 - 6(1)(1)\cos \theta - 8(1)^2 = 0$.
$5 - 6 \cos \theta - 8 = 0$
$-6 \cos \theta = 3$
$\cos \theta = -\frac{3}{6} = -\frac{1}{2}$
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2\pi}{3}$.

(B) Let $\theta$ be the angle between $\overline{a}$ and $\overline{b}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| = 1$ and $|\overline{b}| = 1$.
Given that $\overline{c} = \overline{a} + 2\overline{b}$ and $\overline{d} = 5\overline{a} - 4\overline{b}$ are perpendicular,their dot product is zero:
$\overline{c} \cdot \overline{d} = 0$
$(\overline{a} + 2\overline{b}) \cdot (5\overline{a} - 4\overline{b}) = 0$
$5(\overline{a} \cdot \overline{a}) - 4(\overline{a} \cdot \overline{b}) + 10(\overline{b} \cdot \overline{a}) - 8(\overline{b} \cdot \overline{b}) = 0$
$5|\overline{a}|^2 + 6(\overline{a} \cdot \overline{b}) - 8|\overline{b}|^2 = 0$
Since $|\overline{a}| = 1$ and $|\overline{b}| = 1$,and $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos \theta = \cos \theta$:
$5(1) + 6 \cos \theta - 8(1) = 0$
$6 \cos \theta - 3 = 0$
$6 \cos \theta = 3$
$\cos \theta = \frac{3}{6} = \frac{1}{2}$
$\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

(C) Let $\overline{u} = 3 \overline{a} + 5 \overline{b}$ and $\overline{v} = 5 \overline{a} + 3 \overline{b}$.
$\overline{u} = 3(\hat{i}-2 \hat{j}+3 \hat{k}) + 5(2 \hat{i}+3 \hat{j}-\hat{k}) = (3+10)\hat{i} + (-6+15)\hat{j} + (9-5)\hat{k} = 13 \hat{i} + 9 \hat{j} + 4 \hat{k}$.
$\overline{v} = 5(\hat{i}-2 \hat{j}+3 \hat{k}) + 3(2 \hat{i}+3 \hat{j}-\hat{k}) = (5+6)\hat{i} + (-10+9)\hat{j} + (15-3)\hat{k} = 11 \hat{i} - \hat{j} + 12 \hat{k}$.
The angle $\theta$ between $\overline{u}$ and $\overline{v}$ is given by $\cos \theta = \frac{\overline{u} \cdot \overline{v}}{|\overline{u}| |\overline{v}|}$.
$\overline{u} \cdot \overline{v} = (13)(11) + (9)(-1) + (4)(12) = 143 - 9 + 48 = 182$.
$|\overline{u}| = \sqrt{13^2 + 9^2 + 4^2} = \sqrt{169 + 81 + 16} = \sqrt{266}$.
$|\overline{v}| = \sqrt{11^2 + (-1)^2 + 12^2} = \sqrt{121 + 1 + 144} = \sqrt{266}$.
$\cos \theta = \frac{182}{\sqrt{266} \cdot \sqrt{266}} = \frac{182}{266} = \frac{13}{19}$.
Therefore,$\theta = \cos^{-1}\left(\frac{13}{19}\right)$.

(C) Let $\bar{r}$ be a vector in the plane of $\bar{a}$ and $\bar{b}$. Then,$\bar{r} = \bar{a} + m\bar{b}$ for some scalar $m$.
$\bar{r} = (\hat{i} + 2\hat{j} + \hat{k}) + m(\hat{i} - \hat{j} + \hat{k}) = (1+m)\hat{i} + (2-m)\hat{j} + (1+m)\hat{k}$.
The projection of $\bar{r}$ on $\bar{c}$ is given by $\frac{\bar{r} \cdot \bar{c}}{|\bar{c}|} = \frac{1}{\sqrt{3}}$.
First,calculate $|\bar{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.
Now,$\bar{r} \cdot \bar{c} = (1+m)(1) + (2-m)(1) + (1+m)(-1) = 1+m + 2-m - 1-m = 2-m$.
Thus,$\frac{2-m}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow 2-m = 1 \Rightarrow m = 1$.
Substituting $m=1$ into the expression for $\bar{r}$:
$\bar{r} = (1+1)\hat{i} + (2-1)\hat{j} + (1+1)\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$.
Checking the options,if we consider the projection as $\pm \frac{1}{\sqrt{3}}$,then $2-m = -1 \Rightarrow m = 3$.
For $m=3$,$\bar{r} = (1+3)\hat{i} + (2-3)\hat{j} + (1+3)\hat{k} = 4\hat{i} - \hat{j} + 4\hat{k}$.
This matches option $C$.

(C) First,we find the vectors $\overline{AB}$ and $\overline{CD}$:
$\overline{AB} = (1-2)\hat{i} + (-4-(-3))\hat{j} + (-2-0)\hat{k} = -\hat{i} - \hat{j} - 2\hat{k}$
$\overline{CD} = (7-4)\hat{i} + (0-6)\hat{j} + (10-8)\hat{k} = 3\hat{i} - 6\hat{j} + 2\hat{k}$
The vector projection of $\overline{AB}$ on $\overline{CD}$ is given by the formula:
$\text{Vector Projection} = (\overline{AB} \cdot \hat{CD}) \hat{CD} = (\overline{AB} \cdot \overline{CD}) \frac{\overline{CD}}{|\overline{CD}|^2}$
Calculate the dot product $\overline{AB} \cdot \overline{CD}$:
$\overline{AB} \cdot \overline{CD} = (-1)(3) + (-1)(-6) + (-2)(2) = -3 + 6 - 4 = -1$
Calculate the magnitude squared $|\overline{CD}|^2$:
$|\overline{CD}|^2 = 3^2 + (-6)^2 + 2^2 = 9 + 36 + 4 = 49$
Substitute these values into the formula:
$\text{Vector Projection} = (-1) \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{49} = -\frac{1}{49}(3\hat{i} - 6\hat{j} + 2\hat{k})$

(C) Let $\overline{d} = \overline{a} + \lambda \overline{b}$.
$\overline{d} = (2 \hat{i} + 3 \hat{j} + 2 \hat{k}) + \lambda(2 \hat{i} + \hat{j} - \hat{k})$
$\overline{d} = (2 + 2\lambda) \hat{i} + (3 + \lambda) \hat{j} + (2 - \lambda) \hat{k}$.
Since $\overline{d}$ is perpendicular to $\overline{c}$,their dot product must be zero,i.e.,$\overline{c} \cdot \overline{d} = 0$.
$(\hat{i} + 3 \hat{j}) \cdot [(2 + 2\lambda) \hat{i} + (3 + \lambda) \hat{j} + (2 - \lambda) \hat{k}] = 0$.
$1(2 + 2\lambda) + 3(3 + \lambda) + 0(2 - \lambda) = 0$.
$2 + 2\lambda + 9 + 3\lambda = 0$.
$5\lambda + 11 = 0$.
$\lambda = -\frac{11}{5}$.

(B) Let $\vec{a}$ be the vector joining points $A(-2,0,3)$ and $B(1,4,2)$.
$\vec{a} = (1 - (-2))\hat{i} + (4 - 0)\hat{j} + (2 - 3)\hat{k} = 3\hat{i} + 4\hat{j} - \hat{k}$.
Let $\vec{b}$ be the vector along the line with direction ratios $6, -2, 3$,so $\vec{b} = 6\hat{i} - 2\hat{j} + 3\hat{k}$.
The scalar projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (3)(6) + (4)(-2) + (-1)(3) = 18 - 8 - 3 = 7$.
$|\vec{b}| = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Therefore,the scalar projection is $\frac{7}{7} = 1$.

(A) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Since $(\bar{a}+2 \bar{b})$ and $(5 \bar{a}-4 \bar{b})$ are perpendicular,their dot product is zero:
$(\bar{a}+2 \bar{b}) \cdot (5 \bar{a}-4 \bar{b}) = 0$
$5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{b} \cdot \bar{a}) - 8|\bar{b}|^2 = 0$
Substituting $|\bar{a}|^2 = 1$ and $|\bar{b}|^2 = 1$:
$5(1) + 6(\bar{a} \cdot \bar{b}) - 8(1) = 0$
$6(\bar{a} \cdot \bar{b}) - 3 = 0$
$\bar{a} \cdot \bar{b} = \frac{3}{6} = \frac{1}{2}$
Since $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos \theta$,we have:
$1 \cdot 1 \cdot \cos \theta = \frac{1}{2}$
$\cos \theta = \frac{1}{2}$
$\theta = \frac{\pi}{3}$

(D) Given $\overline{r} \times \overline{a} = \overline{b} \times \overline{a} \implies (\overline{r} - \overline{b}) \times \overline{a} = \overline{0}$. This implies $(\overline{r} - \overline{b}) = k_1 \overline{a}$ for some scalar $k_1$.
Similarly,$\overline{r} \times \overline{b} = \overline{a} \times \overline{b} \implies (\overline{r} - \overline{a}) \times \overline{b} = \overline{0}$. This implies $(\overline{r} - \overline{a}) = k_2 \overline{b}$ for some scalar $k_2$.
From the first equation,$\overline{r} = \overline{b} + k_1 \overline{a} = (2\hat{j} - \hat{k}) + k_1(\hat{i} + \hat{j}) = k_1\hat{i} + (2+k_1)\hat{j} - \hat{k}$.
From the second equation,$\overline{r} = \overline{a} + k_2 \overline{b} = (\hat{i} + \hat{j}) + k_2(2\hat{j} - \hat{k}) = \hat{i} + (1+2k_2)\hat{j} - k_2\hat{k}$.
Comparing the components:
$k_1 = 1$
$2+k_1 = 1+2k_2 \implies 2+1 = 1+2k_2 \implies 2k_2 = 2 \implies k_2 = 1$
$-1 = -k_2 \implies k_2 = 1$.
Substituting $k_1=1$ into $\overline{r} = k_1\hat{i} + (2+k_1)\hat{j} - \hat{k}$,we get $\overline{r} = \hat{i} + 3\hat{j} - \hat{k}$.
The magnitude $|\overline{r}| = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.
Therefore,$\frac{\overline{r}}{|\overline{r}|} = \frac{\hat{i} + 3\hat{j} - \hat{k}}{\sqrt{11}}$.

(B) Given that $\overline{a} \cdot(\overline{b}+\overline{c})+\overline{b} \cdot(\overline{c}+\overline{a})+\overline{c} \cdot(\overline{a}+\overline{b})=0$.
Expanding the terms,we get $\overline{a} \cdot \overline{b} + \overline{a} \cdot \overline{c} + \overline{b} \cdot \overline{c} + \overline{b} \cdot \overline{a} + \overline{c} \cdot \overline{a} + \overline{c} \cdot \overline{b} = 0$.
Since the dot product is commutative,$\overline{a} \cdot \overline{b} = \overline{b} \cdot \overline{a}$,etc.
Thus,$2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a}) = 0$,which implies $\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a} = 0$.
We know that $|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.
Substituting the given values $|\overline{a}|=1, |\overline{b}|=8, |\overline{c}|=4$ and the result from above:
$|\overline{a}+\overline{b}+\overline{c}|^2 = 1^2 + 8^2 + 4^2 + 2(0) = 1 + 64 + 16 = 81$.
Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{81} = 9$.

(A) We are given the expression $E = |(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$.
Expanding the cross product term:
$(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b}) = (\bar{a} \times \bar{a}) + (\bar{a} \times \bar{b}) - (\bar{b} \times \bar{a}) - (\bar{b} \times \bar{b})$.
Since $\bar{a} \times \bar{a} = 0$ and $\bar{b} \times \bar{b} = 0$,and using the property $\bar{b} \times \bar{a} = -(\bar{a} \times \bar{b})$,we get:
$(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b}) = (\bar{a} \times \bar{b}) - (-(\bar{a} \times \bar{b})) = 2(\bar{a} \times \bar{b})$.
Now,substituting this into the expression:
$E = |2(\bar{a} \times \bar{b})|^2 + 4(\bar{a} \cdot \bar{b})^2 = 4|\bar{a} \times \bar{b}|^2 + 4(\bar{a} \cdot \bar{b})^2$.
Factoring out $4$:
$E = 4(|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2)$.
Using Lagrange's identity,$|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2$.
Thus,$E = 4 |\bar{a}|^2 |\bar{b}|^2$.
Given $|\bar{a}|=4$ and $|\bar{b}|=3$:
$E = 4 \times (4)^2 \times (3)^2 = 4 \times 16 \times 9 = 576$.

(B) Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
Given $\overline{r} \times \overline{a} = \overline{b}$,we have:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 2 & 3 & 4 \end{vmatrix} = \hat{i} - 2 \hat{j} + \hat{k}$
$(4y - 3z) \hat{i} - (4x - 2z) \hat{j} + (3x - 2y) \hat{k} = \hat{i} - 2 \hat{j} + \hat{k}$
Comparing coefficients:
$4y - 3z = 1$ $(1)$
$4x - 2z = 2 \Rightarrow 2x - z = 1 \Rightarrow z = 2x - 1$ $(2)$
$3x - 2y = 1 \Rightarrow 2y = 3x - 1 \Rightarrow y = \frac{3x - 1}{2}$ $(3)$
Given $\overline{r} \cdot \overline{c} = 3$,we have $x + y - z = 3$ $(4)$
Substituting $(2)$ and $(3)$ into $(4)$:
$x + \frac{3x - 1}{2} - (2x - 1) = 3$
$2x + 3x - 1 - 4x + 2 = 6$
$x + 1 = 6 \Rightarrow x = 5$
Using $x=5$ in $(2)$ and $(3)$:
$z = 2(5) - 1 = 9$
$y = \frac{3(5) - 1}{2} = 7$
Thus,$\overline{r} = 5 \hat{i} + 7 \hat{j} + 9 \hat{k}$
$|\overline{r}| = \sqrt{5^2 + 7^2 + 9^2} = \sqrt{25 + 49 + 81} = \sqrt{155}$

(B) Given that $\overline{a}, \overline{b}, \overline{c}$ are unit vectors,so $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$.
Given equation: $\overline{a} + 2\overline{c} = -2\overline{b}$.
Squaring both sides:
$|\overline{a} + 2\overline{c}|^2 = |-2\overline{b}|^2$
$|\overline{a}|^2 + 4|\overline{c}|^2 + 4(\overline{a} \cdot \overline{c}) = 4|\overline{b}|^2$
Since $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$,we have:
$1 + 4(1) + 4(\overline{a} \cdot \overline{c}) = 4(1)$
$5 + 4(\overline{a} \cdot \overline{c}) = 4$
$4(\overline{a} \cdot \overline{c}) = -1$
$\overline{a} \cdot \overline{c} = -\frac{1}{4}$
Since $\overline{a} \cdot \overline{c} = |\overline{a}||\overline{c}| \cos \theta = \cos \theta$,we have $\cos \theta = -\frac{1}{4}$.
Then $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (-\frac{1}{4})^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$.
Finally,$|\overline{a} \times \overline{c}| = |\overline{a}||\overline{c}| \sin \theta = (1)(1) \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{4}$.

(B) We are given the expression $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.
Using the distributive property of the dot product,we expand the expression:
$= (a \vec{b}) \cdot (a \vec{b}) - (a \vec{b}) \cdot (b \vec{a}) + (b \vec{a}) \cdot (a \vec{b}) - (b \vec{a}) \cdot (b \vec{a})$
$= a^2 (\vec{b} \cdot \vec{b}) - ab (\vec{b} \cdot \vec{a}) + ab (\vec{a} \cdot \vec{b}) - b^2 (\vec{a} \cdot \vec{a})$
Since the dot product is commutative,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,so the middle terms cancel out:
$= a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2$
Wait,let us re-evaluate the expression given in the prompt: $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.
$= a^2 (\vec{b} \cdot \vec{b}) - b^2 (\vec{a} \cdot \vec{a})$
$= a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2$
If $a$ and $b$ are scalars representing the magnitudes $|\vec{a}|$ and $|\vec{b}|$,then $a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2 = 0$.

(B) We know that $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
Substituting the given values: $35 = \sqrt{26} \times 7 \times \sin \theta$.
$\sin \theta = \frac{35}{7 \sqrt{26}} = \frac{5}{\sqrt{26}}$.
Since $\cos^2 \theta = 1 - \sin^2 \theta$,we have $\cos^2 \theta = 1 - \frac{25}{26} = \frac{1}{26}$.
Thus,$\cos \theta = \frac{1}{\sqrt{26}}$ (assuming $\theta$ is acute).
Now,the dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
$\vec{a} \cdot \vec{b} = \sqrt{26} \times 7 \times \frac{1}{\sqrt{26}} = 7$.

(C) The angle between two vectors $\vec{a}$ and $\vec{b}$ is obtuse if and only if their dot product is negative,i.e.,$\vec{a} \cdot \vec{b} < 0$.
Given $\vec{a} = 2\lambda^2 \hat{i} + 4\lambda \hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + \lambda \hat{k}$.
Calculating the dot product:
$\vec{a} \cdot \vec{b} = (2\lambda^2)(7) + (4\lambda)(-2) + (1)(\lambda) < 0$
$\Rightarrow 14\lambda^2 - 8\lambda + \lambda < 0$
$\Rightarrow 14\lambda^2 - 7\lambda < 0$
$\Rightarrow 7\lambda(2\lambda - 1) < 0$
To solve this inequality,we find the critical points $\lambda = 0$ and $\lambda = \frac{1}{2}$.
Testing the intervals,the expression $7\lambda(2\lambda - 1)$ is negative for $\lambda \in \left(0, \frac{1}{2}\right)$.
Thus,the values of $\lambda$ lie in $\left(0, \frac{1}{2}\right)$.

(A) Given that $\bar{a}, \bar{b}, \bar{c}$ are unit vectors,so $|\bar{a}| = |\bar{b}| = |\bar{c}| = 1$.
We are given $|\bar{a}+\bar{b}+\bar{c}|=1$. Squaring both sides,we get:
$|\bar{a}+\bar{b}+\bar{c}|^2 = 1^2$
$|\bar{a}|^2+|\bar{b}|^2+|\bar{c}|^2+2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 1$
Since $\bar{b} \perp \bar{c}$,we have $\bar{b} \cdot \bar{c} = 0$.
Also,$\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos \alpha = \cos \alpha$ and $\bar{a} \cdot \bar{c} = |\bar{a}||\bar{c}| \cos \beta = \cos \beta$.
Substituting these values into the equation:
$1+1+1+2(\cos \alpha + 0 + \cos \beta) = 1$
$3 + 2(\cos \alpha + \cos \beta) = 1$
$2(\cos \alpha + \cos \beta) = 1 - 3$
$2(\cos \alpha + \cos \beta) = -2$
$\cos \alpha + \cos \beta = -1$

(B) Any vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$ can be expressed as a linear combination: $\vec{V} = x\vec{a} + y\vec{b}$. For simplicity,we can write $\vec{V} = \vec{a} + \lambda\vec{b}$ (assuming the coefficient of $\vec{a}$ is non-zero).
$\vec{V} = (\hat{i}+\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k}) = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1+\lambda)\hat{k}$.
The projection of $\vec{V}$ on $\vec{c}$ is given by $\frac{\vec{V} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.
Given $\vec{c} = \hat{i}-\hat{j}-\hat{k}$,we have $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
So,$\frac{\vec{V} \cdot \vec{c}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \vec{V} \cdot \vec{c} = 1$.
$((1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1+\lambda)\hat{k}) \cdot (\hat{i}-\hat{j}-\hat{k}) = 1$.
$(1+\lambda) - (1-\lambda) - (1+\lambda) = 1$.
$1 + \lambda - 1 + \lambda - 1 - \lambda = 1$.
$\lambda - 1 = 1 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ into the expression for $\vec{V}$:
$\vec{V} = (1+2)\hat{i} + (1-2)\hat{j} + (1+2)\hat{k} = 3\hat{i} - \hat{j} + 3\hat{k}$.

(A) Given $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$.
This implies $\vec{b} \times (\vec{c} - \vec{a}) = \vec{0}$.
Therefore,$\vec{c} - \vec{a} = \lambda \vec{b}$ for some scalar $\lambda$,so $\vec{c} = \vec{a} + \lambda \vec{b}$.
Given $\vec{c} \cdot \vec{a} = 0$,we have $(\vec{a} + \lambda \vec{b}) \cdot \vec{a} = 0$,which means $\vec{a} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.
Calculating the dot products:
$\vec{a} \cdot \vec{a} = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6$.
$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(-2) + (1)(1) = 1 + 2 + 1 = 4$.
Substituting these values: $6 + \lambda(4) = 0 \Rightarrow 4\lambda = -6 \Rightarrow \lambda = -\frac{3}{2}$.
Now,$\vec{c} \cdot \vec{b} = (\vec{a} + \lambda \vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + \lambda (\vec{b} \cdot \vec{b})$.
$\vec{b} \cdot \vec{b} = (1)^2 + (-1)^2 + (1)^2 = 3$.
$\vec{c} \cdot \vec{b} = 4 + (-\frac{3}{2})(3) = 4 - \frac{9}{2} = -\frac{1}{2}$.

(B) Since $\vec{r}$ lies in the plane of $\vec{a}$ and $\vec{b}$,we can write $\vec{r} = t\vec{a} + u\vec{b}$ for some scalars $t$ and $u$.
$\vec{r} = t(\hat{i}+\hat{j}+\hat{k}) + u(\hat{i}-\hat{j}+\hat{k}) = (t+u)\hat{i} + (t-u)\hat{j} + (t+u)\hat{k} \dots (i)$
Given that the projection of $\vec{r}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,we have $\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.
$|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
$\vec{r} \cdot \vec{c} = (t+u)(1) + (t-u)(-1) + (t+u)(-1) = t+u - t+u - t-u = u-t$.
So,$\frac{u-t}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow u-t = 1 \Rightarrow u = t+1$.
Substituting $u = t+1$ into equation $(i)$:
$\vec{r} = (t + t + 1)\hat{i} + (t - (t + 1))\hat{j} + (t + t + 1)\hat{k}$
$\vec{r} = (2t+1)\hat{i} - \hat{j} + (2t+1)\hat{k}$.

(B) Given $\vec{a}+\vec{b} = \lambda \vec{c} \quad \dots(i)$ and $\vec{b}+\vec{c} = \mu \vec{a} \quad \dots(ii)$.
Subtracting $(ii)$ from $(i)$,we get $\vec{a}-\vec{c} = \lambda \vec{c} - \mu \vec{a}$.
Rearranging terms,$(1+\mu)\vec{a} = (1+\lambda)\vec{c}$.
Since two vectors are collinear,let $\vec{a}$ and $\vec{b}$ be collinear,then $\vec{b} = k\vec{a}$.
Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}$,we have $|k\vec{a}| = |\vec{a}| \Rightarrow |k|=1$,so $k=1$ or $k=-1$.
If $k=1$,$\vec{b}=\vec{a}$,then $\vec{a}+\vec{a} = 2\vec{a}$ is collinear with $\vec{c}$,so $\vec{c} = \pm \vec{a}$.
If $\vec{c} = -\vec{a}$,then $\vec{a}+\vec{b}+\vec{c} = \vec{a}+\vec{a}-\vec{a} = \vec{a} \neq 0$. However,the condition $\vec{a}+\vec{b}+\vec{c}=0$ is derived from the collinearity constraints.
Using $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$.
Since $\vec{a}+\vec{b}+\vec{c}=0$,we have $0 = 2+2+2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$.
Thus,$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = -6$.
Therefore,$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = -3$.

(D) The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
Given that $|\vec{a}| = \sqrt{2}$,$|\vec{b}| = \sqrt{2}$,and $\vec{a} \cdot \vec{b} = -1$.
Substituting these values into the formula:
$\cos \theta = \frac{-1}{\sqrt{2} \times \sqrt{2}} = \frac{-1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta$ is $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.

(C) Given: $|\vec{a}| = \sqrt{26}$,$|\vec{b}| = 7$,and $|\vec{a} \times \vec{b}| = 35$.
We know that $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
Substituting the values: $35 = \sqrt{26} \times 7 \times \sin \theta$.
$\sin \theta = \frac{35}{7 \sqrt{26}} = \frac{5}{\sqrt{26}}$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$:
$\cos^2 \theta = 1 - (\frac{5}{\sqrt{26}})^2 = 1 - \frac{25}{26} = \frac{1}{26}$.
Therefore,$\cos \theta = \pm \frac{1}{\sqrt{26}}$.
Now,the dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
$\vec{a} \cdot \vec{b} = \sqrt{26} \times 7 \times (\pm \frac{1}{\sqrt{26}}) = \pm 7$.

(C) Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.
Rearranging the terms,we get $\vec{a}+\vec{b}=-\vec{c}$.
Squaring both sides,we have $|\vec{a}+\vec{b}|^2 = |-\vec{c}|^2$,which implies $|\vec{a}|^2+|\vec{b}|^2+2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.
Substituting the given magnitudes $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$:
$3^2+5^2+2|\vec{a}||\vec{b}| \cos \theta = 7^2$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$9+25+2(3)(5) \cos \theta = 49$.
$34+30 \cos \theta = 49$.
$30 \cos \theta = 49-34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(A) Given $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$.
We know that $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$,so $(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} = -\vec{b} \times (2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Substituting this into the equation,we get $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) + \vec{b} \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.
This implies $(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.
This means the vector $(\vec{a}+\vec{b})$ is parallel to $(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Let $\vec{a}+\vec{b} = \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ for some scalar $\lambda$.
Taking the magnitude on both sides,$|\vec{a}+\vec{b}| = |\lambda| \sqrt{2^2+3^2+4^2} = |\lambda| \sqrt{4+9+16} = |\lambda| \sqrt{29}$.
Given $|\vec{a}+\vec{b}| = \sqrt{29}$,we have $|\lambda| \sqrt{29} = \sqrt{29}$,which gives $\lambda = \pm 1$.
Thus,$\vec{a}+\vec{b} = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Now,calculate the dot product: $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$.
$= \pm((2)(-7) + (3)(2) + (4)(3)) = \pm(-14 + 6 + 12) = \pm 4$.
Therefore,a possible value is $4$.

(C) Given that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,their dot product must be zero:
$(\vec{a}+\lambda \vec{b}) \cdot \vec{c} = 0$
First,calculate $(\vec{a}+\lambda \vec{b})$:
$\vec{a}+\lambda \vec{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k}) = (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$
Now,take the dot product with $\vec{c} = 3 \hat{i} + \hat{j} + 0 \hat{k}$:
$((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (3 \hat{i} + \hat{j} + 0 \hat{k}) = 0$
$3(2-\lambda) + 1(2+2\lambda) + 0(3+\lambda) = 0$
$6 - 3\lambda + 2 + 2\lambda = 0$
$8 - \lambda = 0$
$\lambda = 8$

(D) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Also,$\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
Squaring both sides of the equation,we get:
$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{0}|^2$
$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Substituting the values $|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1$:
$1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$

(C) Given the condition: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS} = \overrightarrow{OQ} \cdot \overrightarrow{OR} + \overrightarrow{OP} \cdot \overrightarrow{OS}$.
Consider the first equality: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS}$.
Rearranging the terms: $\overrightarrow{OP} \cdot \overrightarrow{OQ} - \overrightarrow{OR} \cdot \overrightarrow{OP} = \overrightarrow{OQ} \cdot \overrightarrow{OS} - \overrightarrow{OR} \cdot \overrightarrow{OS}$.
Factoring out the common vectors: $\overrightarrow{OP} \cdot (\overrightarrow{OQ} - \overrightarrow{OR}) = \overrightarrow{OS} \cdot (\overrightarrow{OQ} - \overrightarrow{OR})$.
This implies: $(\overrightarrow{OP} - \overrightarrow{OS}) \cdot (\overrightarrow{OQ} - \overrightarrow{OR}) = 0$.
Since $\overrightarrow{OP} - \overrightarrow{OS} = \overrightarrow{SP}$ and $\overrightarrow{OQ} - \overrightarrow{OR} = \overrightarrow{RQ}$,we have $\overrightarrow{SP} \cdot \overrightarrow{RQ} = 0$,which means $\overrightarrow{PS} \perp \overrightarrow{QR}$.
Similarly,by considering the other parts of the equality,we get $\overrightarrow{QS} \perp \overrightarrow{PR}$ and $\overrightarrow{RS} \perp \overrightarrow{PQ}$.
Since $S$ is the point where the altitudes of the triangle $PQR$ intersect,$S$ is the orthocentre.

(A) We are given $|\vec{a}|=5$,$|\vec{b}|=13$,and $|\vec{a} \times \vec{b}|=25$.
Using the formula $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,we have:
$25 = 5 \times 13 \sin \theta$
$25 = 65 \sin \theta$
$\sin \theta = \frac{25}{65} = \frac{5}{13}$.
Since $\frac{\pi}{2} < \theta < \pi$,the angle $\theta$ lies in the second quadrant,where $\cos \theta$ is negative.
Using $\cos^2 \theta = 1 - \sin^2 \theta$:
$\cos \theta = -\sqrt{1 - (\frac{5}{13})^2} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$.
Now,the dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$:
$\vec{a} \cdot \vec{b} = 5 \times 13 \times (-\frac{12}{13}) = -60$.

(A) First,we find the vectors $\overline{PQ}$ and $\overline{AB}$:
$\overline{PQ} = (3 - (-2)) \hat{i} + (2 - 1) \hat{j} + (5 - 3) \hat{k} = 5 \hat{i} + \hat{j} + 2 \hat{k}$
$\overline{AB} = (7 - 4) \hat{i} + (-5 - (-3)) \hat{j} + (-1 - 5) \hat{k} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k}$
The vector projection of $\overline{PQ}$ on $\overline{AB}$ is given by the formula:
$\text{Vector Projection} = \frac{(\overline{PQ} \cdot \overline{AB}) \overline{AB}}{|\overline{AB}|^2}$
Calculate the dot product $\overline{PQ} \cdot \overline{AB}$:
$\overline{PQ} \cdot \overline{AB} = (5)(3) + (1)(-2) + (2)(-6) = 15 - 2 - 12 = 1$
Calculate the magnitude squared $|\overline{AB}|^2$:
$|\overline{AB}|^2 = 3^2 + (-2)^2 + (-6)^2 = 9 + 4 + 36 = 49$
Thus,the vector projection is:
$\frac{1}{49}(3 \hat{i} - 2 \hat{j} - 6 \hat{k})$

(D) Given that $\bar{a}$ is perpendicular to $\bar{b}+\bar{c}$,$\bar{b}$ is perpendicular to $\bar{c}+\bar{a}$,and $\bar{c}$ is perpendicular to $\bar{a}+\bar{b}$.
Therefore,we have:
$\bar{a} \cdot (\bar{b} + \bar{c}) = 0 \implies \bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} = 0 \quad (1)$
$\bar{b} \cdot (\bar{c} + \bar{a}) = 0 \implies \bar{b} \cdot \bar{c} + \bar{b} \cdot \bar{a} = 0 \quad (2)$
$\bar{c} \cdot (\bar{a} + \bar{b}) = 0 \implies \bar{c} \cdot \bar{a} + \bar{c} \cdot \bar{b} = 0 \quad (3)$
Adding equations $(1), (2),$ and $(3)$,we get:
$2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0 \implies \bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a} = 0$.
Now,consider the magnitude squared of the sum:
$|\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a})$.
Substituting the given values $|\bar{a}|=2, |\bar{b}|=3, |\bar{c}|=4$ and the result $0$ for the dot product sum:
$|\bar{a} + \bar{b} + \bar{c}|^2 = 2^2 + 3^2 + 4^2 + 2(0) = 4 + 9 + 16 = 29$.
Thus,$|\bar{a} + \bar{b} + \bar{c}| = \sqrt{29}$.

(D) Given that $\bar{a} \cdot (\bar{b} + \bar{c}) = 0$,$\bar{b} \cdot (\bar{c} + \bar{a}) = 0$,and $\bar{c} \cdot (\bar{a} + \bar{b}) = 0$.
Expanding these,we get:
$\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} = 0$ . . . $(1)$
$\bar{b} \cdot \bar{c} + \bar{b} \cdot \bar{a} = 0$ . . . $(2)$
$\bar{c} \cdot \bar{a} + \bar{c} \cdot \bar{b} = 0$ . . . $(3)$
Adding equations $(1)$,$(2)$,and $(3)$,we get:
$2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0$.
We know that $|\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a})$.
Substituting the given values $|\bar{a}| = 5, |\bar{b}| = 4, |\bar{c}| = 3$ and the sum of dot products as $0$:
$|\bar{a} + \bar{b} + \bar{c}|^2 = (5)^2 + (4)^2 + (3)^2 + 0 = 25 + 16 + 9 = 50$.

(A) Given vectors are $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$,and $\bar{c}=-3 \hat{i}+\hat{j}+2 \hat{k}$.
First,we find the vector $\bar{a}+\lambda \bar{b}$:
$\bar{a}+\lambda \bar{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k}) = (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$.
Since $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,their dot product must be zero:
$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$.
Substituting the components:
$((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (-3 \hat{i}+\hat{j}+2 \hat{k}) = 0$.
$(2-\lambda)(-3) + (2+2\lambda)(1) + (3+\lambda)(2) = 0$.
$-6 + 3\lambda + 2 + 2\lambda + 6 + 2\lambda = 0$.
$7\lambda + 2 = 0$.
$7\lambda = -2$.
$\lambda = -\frac{2}{7}$.

(C) We are given the identity $|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2$.
Given that $|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = 144$,we have $|\bar{a}|^2 |\bar{b}|^2 = 144$.
Since $|\bar{a}| = 4$,we have $|\bar{a}|^2 = 16$.
Substituting this into the equation,we get $16 |\bar{b}|^2 = 144$.
Dividing both sides by $16$,we get $|\bar{b}|^2 = \frac{144}{16} = 9$.
Taking the square root,we find $|\bar{b}| = 3$.

(C) Given that $|\overline{a} \times \overline{b}| = |\overline{a}| |\overline{b}| \sin \theta = 25$.
Substituting the given values,we get $(5)(13) \sin \theta = 25$.
Thus,$65 \sin \theta = 25$,which implies $\sin \theta = \frac{25}{65} = \frac{5}{13}$.
Since $\frac{\pi}{2} < \theta \leq \pi$,$\theta$ lies in the second quadrant where $\cos \theta$ is negative.
Using $\cos \theta = -\sqrt{1 - \sin^2 \theta}$,we get $\cos \theta = -\sqrt{1 - (\frac{5}{13})^2} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$.
Now,the dot product is $\overline{a} \cdot \overline{b} = |\overline{a}| |\overline{b}| \cos \theta$.
Substituting the values,$\overline{a} \cdot \overline{b} = (5)(13) \left(-\frac{12}{13}\right) = -60$.

(A) Given $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}$,$\bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,and $\bar{c}=\hat{i}+3 \hat{j}+\hat{k}$.
First,calculate $\bar{a}+\lambda \bar{b}$:
$\bar{a}+\lambda \bar{b} = (\hat{i}+2 \hat{j}-3 \hat{k}) + \lambda(3 \hat{i}-\hat{j}+2 \hat{k})$
$= (1+3 \lambda) \hat{i} + (2-\lambda) \hat{j} + (-3+2 \lambda) \hat{k}$.
Since $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,their dot product must be zero:
$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$
$((1+3 \lambda) \hat{i} + (2-\lambda) \hat{j} + (-3+2 \lambda) \hat{k}) \cdot (\hat{i}+3 \hat{j}+\hat{k}) = 0$
$(1+3 \lambda)(1) + (2-\lambda)(3) + (-3+2 \lambda)(1) = 0$
$1 + 3 \lambda + 6 - 3 \lambda - 3 + 2 \lambda = 0$
$4 + 2 \lambda = 0$
$2 \lambda = -4$
$\lambda = -2$.

(A) Given that $\overline{a}+\overline{b}+\overline{c}=\overline{0}$.
We can write this as $\overline{a}+\overline{b}=-\overline{c}$.
Squaring both sides,we get $|\overline{a}+\overline{b}|^2 = |-\overline{c}|^2$,which implies $|\overline{a}+\overline{b}|^2 = |\overline{c}|^2$.
Expanding the left side using the dot product property $|\overline{u}+\overline{v}|^2 = |\overline{u}|^2 + |\overline{v}|^2 + 2|\overline{u}||\overline{v}| \cos \theta$,we have:
$|\overline{a}|^2 + |\overline{b}|^2 + 2|\overline{a}||\overline{b}| \cos \theta = |\overline{c}|^2$.
Substituting the given values $|\overline{a}|=3, |\overline{b}|=5, |\overline{c}|=7$:
$3^2 + 5^2 + 2(3)(5) \cos \theta = 7^2$.
$9 + 25 + 30 \cos \theta = 49$.
$34 + 30 \cos \theta = 49$.
$30 \cos \theta = 49 - 34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,the angle $\theta = \frac{\pi}{3}$.

(D) Two vectors are perpendicular if their dot product is zero.
Given that $(\vec{a}+k \vec{b}) \perp (\vec{a}-k \vec{b})$,we have:
$(\vec{a}+k \vec{b}) \cdot (\vec{a}-k \vec{b}) = 0$
$|\vec{a}|^2 - k^2 |\vec{b}|^2 = 0$
Substituting the given values $|\vec{a}|=4$ and $|\vec{b}|=5$:
$(4)^2 - k^2 (5)^2 = 0$
$16 - 25k^2 = 0$
$25k^2 = 16$
$k^2 = \frac{16}{25}$
$k = \pm \frac{4}{5}$

(C) Given that $\hat{a}$ is a unit vector,we have $|\hat{a}| = 1$.
Using the property of the dot product $( \bar{x} - \hat{a} ) \cdot ( \bar{x} + \hat{a} ) = |\bar{x}|^2 - |\hat{a}|^2$.
Substituting the given value: $|\bar{x}|^2 - |\hat{a}|^2 = 8$.
Since $|\hat{a}| = 1$,we have $|\hat{x}|^2 - 1^2 = 8$.
$|\bar{x}|^2 - 1 = 8$.
$|\bar{x}|^2 = 9$.
Since the magnitude of a vector must be non-negative,$|\bar{x}| = 3$.

(A) Given that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,their dot product must be zero: $(\vec{a}+\lambda\vec{b}) \cdot \vec{c} = 0$.
First,calculate $\vec{a}+\lambda\vec{b}$:
$\vec{a}+\lambda\vec{b} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(-\hat{i}+2\hat{j}+\hat{k}) = (1-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}$.
Now,take the dot product with $\vec{c} = 3\hat{i}+\hat{j}$:
$((1-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}) \cdot (3\hat{i}+\hat{j}) = 0$.
$(1-\lambda)(3) + (2+2\lambda)(1) + (3+\lambda)(0) = 0$.
$3 - 3\lambda + 2 + 2\lambda = 0$.
$5 - \lambda = 0$.
Therefore,$\lambda = 5$.

(C) The formula for the projection of vector $\bar{a}$ on vector $\bar{b}$ is given by $\frac{\bar{a} \cdot \bar{b}}{|\bar{b}|}$.
First,calculate the dot product $\bar{a} \cdot \bar{b} = (1)(2) + (-2)(-1) + (1)(1) = 2 + 2 + 1 = 5$.
Next,calculate the magnitude of vector $\bar{b}$,which is $|\bar{b}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Finally,the projection is $\frac{5}{\sqrt{6}}$.

(B) Given vectors are $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$.
Since the angle $\theta$ between $\bar{a}$ and $\bar{b}$ is obtuse,we have $\cos \theta < 0$.
We know that $\cos \theta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}| |\bar{b}|}$.
Since $|\bar{a}| > 0$ and $|\bar{b}| > 0$,the condition $\cos \theta < 0$ implies $\bar{a} \cdot \bar{b} < 0$.
Calculating the dot product: $\bar{a} \cdot \bar{b} = (2 \lambda^2)(7) + (4 \lambda)(-2) + (1)(\lambda) = 14 \lambda^2 - 8 \lambda + \lambda = 14 \lambda^2 - 7 \lambda$.
Setting the dot product to be less than zero: $14 \lambda^2 - 7 \lambda < 0$.
$7 \lambda (2 \lambda - 1) < 0$.
This inequality holds when $\lambda$ lies between the roots $0$ and $\frac{1}{2}$.
Therefore,$\lambda \in \left(0, \frac{1}{2}\right)$.

(D) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4\hat{i}-2\hat{j}$,$\vec{b} = \hat{i}+4\hat{j}-3\hat{k}$,and $\vec{c} = -\hat{i}+5\hat{j}+\hat{k}$.
We need to find $\angle ABC$,which is the angle between vectors $\vec{BA}$ and $\vec{BC}$.
First,calculate $\vec{BA} = \vec{a} - \vec{b} = (4-1)\hat{i} + (-2-4)\hat{j} + (0-(-3))\hat{k} = 3\hat{i} - 6\hat{j} + 3\hat{k}$.
Next,calculate $\vec{BC} = \vec{c} - \vec{b} = (-1-1)\hat{i} + (5-4)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} + \hat{j} + 4\hat{k}$.
Now,find the dot product $\vec{BA} \cdot \vec{BC} = (3)(-2) + (-6)(1) + (3)(4) = -6 - 6 + 12 = 0$.
Since the dot product of $\vec{BA}$ and $\vec{BC}$ is $0$,the vectors are perpendicular to each other.
Therefore,$\angle ABC = \frac{\pi}{2}$.

(D) Using the scalar triple product property $(\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C} = \overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C})$:
$(\overrightarrow{a} \times \hat{j}) \cdot (2 \hat{j} - 3 \hat{k}) = \overrightarrow{a} \cdot \{\hat{j} \times (2 \hat{j} - 3 \hat{k})\}$
$= \overrightarrow{a} \cdot \{2(\hat{j} \times \hat{j}) - 3(\hat{j} \times \hat{k})\}$
Since $\hat{j} \times \hat{j} = 0$ and $\hat{j} \times \hat{k} = \hat{i}$:
$= \overrightarrow{a} \cdot \{2(0) - 3(\hat{i})\}$
$= \overrightarrow{a} \cdot (-3 \hat{i})$
$= -3(\overrightarrow{a} \cdot \hat{i})$
Given $\overrightarrow{a} \cdot \hat{i} = 4$,we substitute this value:
$= -3(4) = -12$

(C) The resultant force $\overrightarrow{F}$ is the sum of the individual forces:
$\overrightarrow{F} = (2 \hat{i}-5 \hat{j}+6 \hat{k}) + (-\hat{i}+2 \hat{j}-\hat{k}) = \hat{i}-3 \hat{j}+5 \hat{k}$
The displacement vector $\overrightarrow{d}$ is given by $\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}$:
$\overrightarrow{d} = (6 \hat{i}+\hat{j}-3 \hat{k}) - (4 \hat{i}-3 \hat{j}-2 \hat{k}) = 2 \hat{i}+4 \hat{j}-\hat{k}$
The work done $W$ is the dot product of the force and displacement vectors:
$W = \overrightarrow{F} \cdot \overrightarrow{d} = (\hat{i}-3 \hat{j}+5 \hat{k}) \cdot (2 \hat{i}+4 \hat{j}-\hat{k})$
$W = (1)(2) + (-3)(4) + (5)(-1) = 2 - 12 - 5 = -15 \text{ unit}$

(D) Given,$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$
$\therefore \overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b})$
Squaring both sides,we get:
$|\overrightarrow{c}|^2 = (\overrightarrow{a}+\overrightarrow{b}) \cdot (\overrightarrow{a}+\overrightarrow{b})$
$|\overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b})$
Since $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos \theta$,where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$:
$|\overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2|\overrightarrow{a}||\overrightarrow{b}| \cos \theta$
Substituting the given values:
$7^2 = 3^2 + 5^2 + 2(3)(5) \cos \theta$
$49 = 9 + 25 + 30 \cos \theta$
$49 = 34 + 30 \cos \theta$
$15 = 30 \cos \theta$
$\cos \theta = \frac{15}{30} = \frac{1}{2}$
Therefore,$\theta = \frac{\pi}{3}$.