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

(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Since $a - \sqrt{2}b$ is a unit vector,its magnitude is $1$,so $|a - \sqrt{2}b| = 1$.
Squaring both sides,we get $|a - \sqrt{2}b|^2 = 1^2$.
Using the property $|u - v|^2 = |u|^2 + |v|^2 - 2(u \cdot v)$,we have:
$|a|^2 + |\sqrt{2}b|^2 - 2(a \cdot \sqrt{2}b) = 1$.
Substituting the values $|a| = 1$ and $|b| = 1$:
$1^2 + 2|b|^2 - 2\sqrt{2}(a \cdot b) = 1$.
$1 + 2(1) - 2\sqrt{2}(a \cdot b) = 1$.
$3 - 2\sqrt{2}(a \cdot b) = 1$.
$2\sqrt{2}(a \cdot b) = 2$.
$a \cdot b = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $a \cdot b = |a||b|\cos\theta$,we have $1 \cdot 1 \cdot \cos\theta = \frac{1}{\sqrt{2}}$.
$\cos\theta = \frac{1}{\sqrt{2}}$,which implies $\theta = \frac{\pi}{4}$.

(D) Let $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
The dot product of two vectors is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,where $\theta$ is the angle between them.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-1)(2) + (1)(1) = 1 - 2 + 1 = 0$.
Next,calculate the magnitudes: $|\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$.
Substitute these values into the formula: $0 = \sqrt{3} \cdot \sqrt{6} \cdot \cos \theta$.
Since $\sqrt{3} \cdot \sqrt{6} \neq 0$,we must have $\cos \theta = 0$.
Therefore,$\theta = \cos^{-1}(0) = \frac{\pi}{2}$.

(D) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4i - 2j + 0k$,$\vec{b} = i + 4j - 3k$,and $\vec{c} = -i + 5j + k$ respectively.
First,we find the vectors $\vec{BA}$ and $\vec{BC}$:
$\vec{BA} = \vec{a} - \vec{b} = (4-1)i + (-2-4)j + (0-(-3))k = 3i - 6j + 3k$
$\vec{BC} = \vec{c} - \vec{b} = (-1-1)i + (5-4)j + (1-(-3))k = -2i + j + 4k$
Now,calculate the dot product $\vec{BA} \cdot \vec{BC}$:
$\vec{BA} \cdot \vec{BC} = (3)(-2) + (-6)(1) + (3)(4) = -6 - 6 + 12 = 0$
Since the dot product of the two vectors is $0$,the vectors $\vec{BA}$ and $\vec{BC}$ are perpendicular to each other.
Therefore,$\angle ABC = 90^\circ$ or $\pi / 2$ radians.

(D) Let $\vec{a} = x\hat{i} - 3\hat{j} - \hat{k}$ and $\vec{b} = 2x\hat{i} + x\hat{j} - \hat{k}$.
Condition $1$: The angle between $\vec{a}$ and $\vec{b}$ is acute,so $\vec{a} \cdot \vec{b} > 0$.
$\vec{a} \cdot \vec{b} = (x)(2x) + (-3)(x) + (-1)(-1) = 2x^2 - 3x + 1 > 0$.
Solving $2x^2 - 3x + 1 > 0$,we get $(2x - 1)(x - 1) > 0$,which implies $x < 1/2$ or $x > 1$.
Condition $2$: The angle between $\vec{b}$ and the $y$-axis (unit vector $\hat{j}$) is obtuse,so $\vec{b} \cdot \hat{j} < 0$.
$\vec{b} \cdot \hat{j} = (2x\hat{i} + x\hat{j} - \hat{k}) \cdot (0\hat{i} + 1\hat{j} + 0\hat{k}) = x < 0$.
Combining both conditions: $x < 0$ and ($x < 1/2$ or $x > 1$).
The intersection is $x < 0$.
Since none of the options match $x < 0$,the correct answer is $D$.

(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Also,$|a - b| = 1$.
Squaring both sides of $|a - b| = 1$,we get $|a - b|^2 = 1^2$.
Using the property $|x|^2 = x \cdot x$,we have $(a - b) \cdot (a - b) = 1$.
Expanding the dot product,we get $a \cdot a - a \cdot b - b \cdot a + b \cdot b = 1$.
Since $a \cdot a = |a|^2 = 1$ and $b \cdot b = |b|^2 = 1$,the equation becomes $1 - 2(a \cdot b) + 1 = 1$.
$2 - 2(a \cdot b) = 1$.
$2(a \cdot b) = 1$,which implies $a \cdot b = \frac{1}{2}$.
We know that $a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between $a$ and $b$.
Substituting the values,$1 \times 1 \times \cos \theta = \frac{1}{2}$.
$\cos \theta = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$ or $60^\circ$.

(C) The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta,$ where $\theta$ is the angle between the vectors,$0 \le \theta \le \pi.$
Since the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are always non-negative,the sign of $\vec{a} \cdot \vec{b}$ depends on $\cos \theta.$
For $\vec{a} \cdot \vec{b} \ge 0,$ we must have $\cos \theta \ge 0.$
In the interval $0 \le \theta \le \pi,$ $\cos \theta \ge 0$ holds when $0 \le \theta \le \frac{\pi}{2}.$
Therefore,the correct option is $C.$

(C) Given vectors are $a = i + 2j - 3k$ and $b = 3i - j + 2k.$
First,calculate the sum vector $a + b$:
$a + b = (i + 3i) + (2j - j) + (-3k + 2k) = 4i + j - k.$
Next,calculate the difference vector $a - b$:
$a - b = (i - 3i) + (2j - (-j)) + (-3k - 2k) = -2i + 3j - 5k.$
Now,find the dot product of $(a + b)$ and $(a - b)$:
$(a + b) \cdot (a - b) = (4)(-2) + (1)(3) + (-1)(-5) = -8 + 3 + 5 = 0.$
Since the dot product of the two vectors is $0$,the vectors are perpendicular to each other.
Therefore,the angle between them is $90^o$.

(B) For the angle between vectors $a$ and $b$ to be acute, their dot product must be positive: $a \cdot b > 0$.
$(-3i + xj + k) \cdot (xi + 2xj + k) > 0$
$-3x + 2x^2 + 1 > 0$
$2x^2 - 3x + 1 > 0$
$(2x - 1)(x - 1) > 0$
This inequality holds when $x < 1/2$ or $x > 1$.
For the angle between vector $b = xi + 2xj + k$ and the $x$-axis (represented by unit vector $i$) to be obtuse (between $\pi/2$ and $\pi$), the dot product $b \cdot i$ must be negative:
$b \cdot i < 0$
$(xi + 2xj + k) \cdot i < 0$
$x < 0$.
Combining the conditions $x < 1/2$ or $x > 1$ and $x < 0$, we get $x < 0$.

(C) Let $\vec{a} = 2i + 6j + 3k$ and $\vec{b} = 12i - 4j + 3k$.
The angle $\theta$ between two vectors is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (2)(12) + (6)(-4) + (3)(3) = 24 - 24 + 9 = 9$.
Next,calculate the magnitudes: $|\vec{a}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.
$|\vec{b}| = \sqrt{12^2 + (-4)^2 + 3^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.
Thus,$\cos \theta = \frac{9}{7 \times 13} = \frac{9}{91}$.
Therefore,$\theta = \cos^{-1}\left(\frac{9}{91}\right)$.

(D) Let $\vec{u} = i + 0j + k$ and $\vec{v} = i - j + ak$.
The dot product is given by $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$.
Here,$\vec{u} \cdot \vec{v} = (1)(1) + (0)(-1) + (1)(a) = 1 + a$.
The magnitudes are $|\vec{u}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$ and $|\vec{v}| = \sqrt{1^2 + (-1)^2 + a^2} = \sqrt{2 + a^2}$.
Given $\theta = \pi / 3$,we have $\cos(\pi / 3) = 1/2$.
Substituting these into the formula: $1 + a = \sqrt{2} \cdot \sqrt{2 + a^2} \cdot (1/2)$.
Multiplying by $2$: $2(1 + a) = \sqrt{2(2 + a^2)}$.
Squaring both sides: $4(1 + 2a + a^2) = 2(2 + a^2)$.
$4 + 8a + 4a^2 = 4 + 2a^2$.
$2a^2 + 8a = 0$.
$2a(a + 4) = 0$.
Thus,$a = 0$ or $a = -4$.
Checking the options,$a = 0$ is provided.

(C) Given,$a + b + c = 0 \Rightarrow a + b = -c$.
Squaring both sides,we get $|a + b|^2 = |-c|^2$.
Using the property $|u|^2 = u \cdot u$,we have $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Since $a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between $a$ and $b$,we have $|a|^2 + |b|^2 + 2|a||b| \cos \theta = |c|^2$.
Substituting the given values $|a| = 3, |b| = 5, |c| = 7$:
$3^2 + 5^2 + 2(3)(5) \cos \theta = 7^2$.
$9 + 25 + 30 \cos \theta = 49$.
$34 + 30 \cos \theta = 49$.
$30 \cos \theta = 49 - 34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Therefore,$\theta = 60^\circ$.

(D) Given that $a, b,$ and $c$ are unit vectors,so $|a| = |b| = |c| = 1.$
The given condition is $a + b - c = 0,$ which implies $a + b = c.$
Taking the dot product of both sides with themselves:
$(a + b) \cdot (a + b) = c \cdot c$
Expanding the dot product:
$a \cdot a + b \cdot b + 2(a \cdot b) = |c|^2$
Since $a \cdot a = |a|^2 = 1,$ $b \cdot b = |b|^2 = 1,$ and $c \cdot c = |c|^2 = 1,$
$1 + 1 + 2|a||b|\cos \theta = 1,$
where $\theta$ is the angle between $a$ and $b.$
$2 + 2(1)(1)\cos \theta = 1$
$2\cos \theta = 1 - 2$
$2\cos \theta = -1$
$\cos \theta = -\frac{1}{2}$
Since $\cos \theta = -\frac{1}{2},$ the angle $\theta = \frac{2\pi}{3}.$

(A) Let $a = 2i + 3j + k$ and $b = 2i - j - k$.
The cosine of the angle $\theta$ between two vectors $a$ and $b$ is given by the formula:
$\cos \theta = \frac{a \cdot b}{|a| |b|}$.
First,calculate the dot product $a \cdot b$:
$a \cdot b = (2)(2) + (3)(-1) + (1)(-1) = 4 - 3 - 1 = 0$.
Next,calculate the magnitudes $|a|$ and $|b|$:
$|a| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
$|b| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Substitute these values into the formula:
$\cos \theta = \frac{0}{\sqrt{14} \cdot \sqrt{6}} = 0$.
Since $\cos \theta = 0$,the angle $\theta = \frac{\pi}{2}$.

(A) The formula for the cosine of the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4$.
Next,calculate the magnitudes of the vectors:
$|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
$|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
Finally,substitute these values into the formula:
$\cos \theta = \frac{4}{3 \times 7} = \frac{4}{21}$.

(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Since $(a + 2b)$ and $(5a - 4b)$ are perpendicular,their dot product is zero:
$(a + 2b) \cdot (5a - 4b) = 0$
Expanding the dot product:
$5(a \cdot a) - 4(a \cdot b) + 10(b \cdot a) - 8(b \cdot b) = 0$
Since $a \cdot a = |a|^2 = 1$ and $b \cdot b = |b|^2 = 1$,and $a \cdot b = b \cdot a$:
$5(1) + 6(a \cdot b) - 8(1) = 0$
$5 + 6(a \cdot b) - 8 = 0$
$6(a \cdot b) - 3 = 0$
$6(a \cdot b) = 3$
$a \cdot b = \frac{3}{6} = \frac{1}{2}$
Using the formula $a \cdot b = |a||b| \cos \theta$:
$(1)(1) \cos \theta = \frac{1}{2}$
$\cos \theta = \frac{1}{2}$
$\theta = 60^o$.

(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
We know that $|a - b|^2 = |a|^2 + |b|^2 - 2|a||b|\cos\theta$.
Substituting the values,we get $|a - b|^2 = 1^2 + 1^2 - 2(1)(1)\cos\theta = 2 - 2\cos\theta$.
Using the trigonometric identity $1 - \cos\theta = 2\sin^2(\theta/2)$,we have $|a - b|^2 = 2(2\sin^2(\theta/2)) = 4\sin^2(\theta/2)$.
Taking the square root on both sides,we get $|a - b| = 2\sin(\theta/2)$.
Therefore,$\sin(\theta/2) = \frac{1}{2}|a - b|$.

(A) Given vectors are $a = (1, 1, 4)$ and $b = (1, -1, 4)$.
First,calculate $a + b$:
$a + b = (1+1, 1-1, 4+4) = (2, 0, 8)$.
Next,calculate $a - b$:
$a - b = (1-1, 1-(-1), 4-4) = (0, 2, 0)$.
Now,find the dot product of $(a + b)$ and $(a - b)$:
$(a + b) \cdot (a - b) = (2)(0) + (0)(2) + (8)(0) = 0 + 0 + 0 = 0$.
Since the dot product of the two vectors is $0$,the vectors are perpendicular to each other.
Therefore,the angle $\theta$ between them is $90^o$.

(B) Given that $|a + b| = |a - b|$.
Squaring both sides,we get $|a + b|^2 = |a - b|^2$.
Using the property $|x|^2 = x \cdot x$,we have $(a + b) \cdot (a + b) = (a - b) \cdot (a - b)$.
Expanding the dot product,we get $a \cdot a + 2(a \cdot b) + b \cdot b = a \cdot a - 2(a \cdot b) + b \cdot b$.
Subtracting $a \cdot a + b \cdot b$ from both sides,we get $2(a \cdot b) = -2(a \cdot b)$.
This implies $4(a \cdot b) = 0$,which means $a \cdot b = 0$.
Since the dot product of two non-zero vectors is zero,the vectors $a$ and $b$ are perpendicular to each other.

(D) Two vectors $\vec{u} = 2\hat{i} + a\hat{j} + \hat{k}$ and $\vec{v} = 2\hat{i} - \hat{j} - \hat{k}$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{u} \cdot \vec{v} = 0.$
Calculating the dot product:
$(2)(2) + (a)(-1) + (1)(-1) = 0$
$4 - a - 1 = 0$
$3 - a = 0$
$a = 3.$
Therefore,the correct option is $D$.

(D) Given vectors are $a = 2i + 2j + 3k$,$b = -i + 2j + k$,and $c = 3i + j$.
First,calculate the vector $a + tb$:
$a + tb = (2i + 2j + 3k) + t(-i + 2j + k) = (2 - t)i + (2 + 2t)j + (3 + t)k$.
Since $a + tb$ is perpendicular to $c$,their dot product must be zero:
$(a + tb) \cdot c = 0$.
Substituting the components:
$((2 - t)i + (2 + 2t)j + (3 + t)k) \cdot (3i + j + 0k) = 0$.
$3(2 - t) + 1(2 + 2t) + 0(3 + t) = 0$.
$6 - 3t + 2 + 2t = 0$.
$8 - t = 0$.
$t = 8$.

(C) Two vectors $\vec{a} = a_1i + a_2j + a_3k$ and $\vec{b} = b_1i + b_2j + b_3k$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
Given vectors are $\vec{a} = 2i + j - k$ and $\vec{b} = i - 4j + \lambda k$.
Calculating the dot product: $\vec{a} \cdot \vec{b} = (2)(1) + (1)(-4) + (-1)(\lambda) = 0$.
$2 - 4 - \lambda = 0$.
$-2 - \lambda = 0$.
$\lambda = -2$.

(B) Two vectors $\vec{A} = A_1\,i + A_2\,j + A_3\,k$ and $\vec{B} = B_1\,i + B_2\,j + B_3\,k$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.
Given vectors are $\vec{A} = 2\,i + 3\,j - 4\,k$ and $\vec{B} = a\,i + b\,j + c\,k$.
The condition for perpendicularity is $(2)(a) + (3)(b) + (-4)(c) = 0$,which simplifies to $2a + 3b - 4c = 0$.
Checking option $(b)$: $2(4) + 3(4) - 4(5) = 8 + 12 - 20 = 20 - 20 = 0$.
Since the dot product is $0$,the vectors are perpendicular for the values given in option $(b)$.

(B) Let the required unit vector in the $xy$-plane be $\vec{r} = xi + yj$.
Since $\vec{r}$ is a unit vector, we have $x^2 + y^2 = 1$.
Given that $\vec{r}$ is perpendicular to $\vec{a} = 4i - 3j + k$, their dot product must be zero:
$(xi + yj) \cdot (4i - 3j + k) = 0$
$4x - 3y = 0 \Rightarrow 4x = 3y \Rightarrow x = \frac{3}{4}y$.
Substituting $x = \frac{3}{4}y$ into $x^2 + y^2 = 1$:
$(\frac{3}{4}y)^2 + y^2 = 1$
$\frac{9}{16}y^2 + y^2 = 1 \Rightarrow \frac{25}{16}y^2 = 1$
$y^2 = \frac{16}{25} \Rightarrow y = \pm \frac{4}{5}$.
If $y = \frac{4}{5}$, then $x = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$.
Thus, the vector is $\frac{3}{5}i + \frac{4}{5}j = \frac{1}{5}(3i + 4j)$.
Comparing this with the given options, the correct option is $B$.

(C) Given the equation $l\vec{a} + m\vec{b} + n\vec{c} = \vec{0}.$
Taking the dot product of the equation with $\vec{a}$ on both sides:
$(l\vec{a} + m\vec{b} + n\vec{c}) \cdot \vec{a} = \vec{0} \cdot \vec{a}$
$l(\vec{a} \cdot \vec{a}) + m(\vec{b} \cdot \vec{a}) + n(\vec{c} \cdot \vec{a}) = 0$
Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular,$\vec{a} \cdot \vec{b} = 0, \vec{b} \cdot \vec{c} = 0,$ and $\vec{c} \cdot \vec{a} = 0.$
Thus,$l|\vec{a}|^2 + 0 + 0 = 0.$
Since $\vec{a}$ is a non-zero vector,$|\vec{a}|^2 \neq 0,$ which implies $l = 0.$
Similarly,taking the dot product with $\vec{b}$ and $\vec{c}$ respectively,we get $m = 0$ and $n = 0.$
Therefore,$l = m = n = 0.$

(A) Two vectors $\vec{u} = a\,i - 2j + 3k$ and $\vec{v} = 3i + 6j - 5k$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{u} \cdot \vec{v} = 0$.
Calculating the dot product: $(a)(3) + (-2)(6) + (3)(-5) = 0$.
$3a - 12 - 15 = 0$.
$3a - 27 = 0$.
$3a = 27$.
$a = 9$.

(A) Two vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
Let $\vec{a} = 2\lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 0\hat{i} + 2\hat{j} + \hat{k}$.
Calculating the dot product: $\vec{a} \cdot \vec{b} = (2\lambda)(0) + (1)(2) + (-1)(1) = 0 + 2 - 1 = 1$.
Since the dot product is $1$,which is not equal to $0$ for any value of $\lambda$,there is no value of $\lambda$ for which these vectors are perpendicular.

(C) Two vectors $\vec{A} = ai + bj + ck$ and $\vec{B} = pi + qj + rk$ are perpendicular if and only if their dot product is zero.
The dot product is given by $\vec{A} \cdot \vec{B} = (ai + bj + ck) \cdot (pi + qj + rk)$.
Calculating the dot product, we get $\vec{A} \cdot \vec{B} = a(p) + b(q) + c(r) = ap + bq + cr$.
Since the vectors are perpendicular, $\vec{A} \cdot \vec{B} = 0$.
Therefore, $ap + bq + cr = 0$.

(C) Two vectors $a$ and $b$ are perpendicular if and only if their dot product is zero,i.e.,$a \cdot b = 0$.
Given $a = 2i + 4j + 2k$ and $b = 8i - 3j + \lambda k$.
The dot product is calculated as:
$a \cdot b = (2)(8) + (4)(-3) + (2)(\lambda) = 0$
$16 - 12 + 2\lambda = 0$
$4 + 2\lambda = 0$
$2\lambda = -4$
$\lambda = -2$
Therefore,the value of $\lambda$ is $-2$.

(D) Two vectors $\vec{u} = a\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{v} = -\hat{i} + 5\hat{j} + a\hat{k}$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{u} \cdot \vec{v} = 0$.
Calculating the dot product:
$(a)(-1) + (2)(5) + (3)(a) = 0$
$-a + 10 + 3a = 0$
$2a + 10 = 0$
$2a = -10$
$a = -5$
Therefore,the correct value is $a = -5$.

(A) Two vectors are perpendicular if and only if their dot product is equal to zero.
Let $\vec{u} = ai + 6j - k$ and $\vec{v} = 7i - 3j + 17k$.
Since $\vec{u} \perp \vec{v}$,we have $\vec{u} \cdot \vec{v} = 0$.
$(ai + 6j - k) \cdot (7i - 3j + 17k) = 0$
$(a)(7) + (6)(-3) + (-1)(17) = 0$
$7a - 18 - 17 = 0$
$7a - 35 = 0$
$7a = 35$
$a = 5$.

(C) Two vectors $\vec{a} = 4\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} + m\hat{j} + 2\hat{k}$ are at a right angle,which means they are perpendicular to each other.
For perpendicular vectors,their dot product must be zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
$(4\hat{i} + \hat{j} - \hat{k}) \cdot (3\hat{i} + m\hat{j} + 2\hat{k}) = 0$
$(4)(3) + (1)(m) + (-1)(2) = 0$
$12 + m - 2 = 0$
$10 + m = 0$
$m = -10$.

(C) Let $\vec{a} = 3i + \lambda j + k$ and $\vec{b} = 2i - j + 8k$.
Since the vectors $\vec{a}$ and $\vec{b}$ are perpendicular,their dot product must be zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
$(3i + \lambda j + k) \cdot (2i - j + 8k) = 0$.
Calculating the dot product by multiplying corresponding components:
$(3)(2) + (\lambda)(-1) + (1)(8) = 0$.
$6 - \lambda + 8 = 0$.
$14 - \lambda = 0$.
Therefore,$\lambda = 14$.

(D) The projection (component) of vector $b$ along vector $a$ is given by the formula: $\text{proj}_a b = (b \cdot \hat{a})\hat{a}$.
Since the unit vector $\hat{a} = \frac{a}{|a|}$,we substitute this into the formula:
$\text{proj}_a b = \left( b \cdot \frac{a}{|a|} \right) \frac{a}{|a|}$.
This simplifies to $\frac{(b \cdot a)a}{|a|^2}$.
Since $|a|^2 = a \cdot a$,the expression becomes $\frac{(a \cdot b)a}{a \cdot a}$.

(B) The component of vector $a$ along $b$ is given by the formula: $\text{proj}_{b} a = \frac{(a \cdot b)b}{|b|^2}$.
Given $a = 4i + 6j$ and $b = 3j + 4k$.
First,calculate the dot product $a \cdot b = (4i + 6j) \cdot (0i + 3j + 4k) = (4 \times 0) + (6 \times 3) + (0 \times 4) = 0 + 18 + 0 = 18$.
Next,calculate the square of the magnitude of $b$: $|b|^2 = 3^2 + 4^2 = 9 + 16 = 25$.
Substituting these values into the formula: $\text{proj}_{b} a = \frac{18}{25}(3j + 4k)$.

(B) Let $\vec{a} = \vec{i} + \vec{j}$ and $\vec{b} = \vec{j} + \vec{k}$.
The component of vector $\vec{a}$ along vector $\vec{b}$ is given by the formula: $\left( \vec{a} \cdot \hat{b} \right) \hat{b}$,where $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$.
First,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
So,$\hat{b} = \frac{\vec{j} + \vec{k}}{\sqrt{2}}$.
Now,calculate the dot product $\vec{a} \cdot \hat{b} = (\vec{i} + \vec{j}) \cdot \left( \frac{\vec{j} + \vec{k}}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} (\vec{i} \cdot \vec{j} + \vec{i} \cdot \vec{k} + \vec{j} \cdot \vec{j} + \vec{j} \cdot \vec{k}) = \frac{1}{\sqrt{2}} (0 + 0 + 1 + 0) = \frac{1}{\sqrt{2}}$.
Finally,the component is $\left( \frac{1}{\sqrt{2}} \right) \left( \frac{\vec{j} + \vec{k}}{\sqrt{2}} \right) = \frac{\vec{j} + \vec{k}}{2}$.

(B) The projection of a vector $\vec{a}$ on a vector $\vec{b}$ is given by the formula: $\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Given $\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (-2)(3) = 2 + 6 - 6 = 2$.
Next,calculate the magnitude of vector $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Therefore,the projection is $\frac{2}{\sqrt{14}}$.

(B) The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
The projection of vector $\vec{b}$ on vector $\vec{a}$ is given by $\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$.
Therefore,the required ratio is $\frac{\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}}{\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}} = \frac{|\vec{a}|}{|\vec{b}|}$.
Given $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$,we have $|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Given $\vec{b} = -2\hat{i} + 2\hat{j} - \hat{k}$,we have $|\vec{b}| = \sqrt{(-2)^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,the required ratio is $\frac{7}{3}$.

(C) The projection of a vector $\vec{a}$ along a vector $\vec{b}$ is defined as the scalar component of $\vec{a}$ in the direction of $\vec{b}$.
Mathematically,this is given by the dot product of $\vec{a}$ and the unit vector in the direction of $\vec{b}$.
Let $\hat{b}$ be the unit vector along $\vec{b}$,where $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$.
The projection is $\vec{a} \cdot \hat{b} = \vec{a} \cdot \frac{\vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Therefore,the correct option is $C$.

(A) The projection of vector $b$ on vector $a$ is given by the formula: $\text{Projection} = \frac{a \cdot b}{|a|}$.
Given vectors are $a = 2i + j + 2k$ and $b = 5i - 3j + k$.
First,calculate the dot product $a \cdot b = (2)(5) + (1)(-3) + (2)(1) = 10 - 3 + 2 = 9$.
Next,calculate the magnitude of vector $a$: $|a| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
Finally,the projection is $\frac{a \cdot b}{|a|} = \frac{9}{3} = 3$.

(B) The projection of a vector $\vec{a}$ on a vector $\vec{b}$ is given by the formula: $\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Given $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(4) + (-2)(-4) + (1)(7) = 4 + 8 + 7 = 19$.
Next,calculate the magnitude of vector $\vec{b}$:
$|\vec{b}| = \sqrt{4^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$.
Therefore,the projection is $\frac{19}{9}$.

(A) The displacement vector $\vec{d}$ is given by $\vec{AB} = \vec{OB} - \vec{OA}$.
$\vec{AB} = (4i - 3j - 2k) - (6i + j - 3k) = (4-6)i + (-3-1)j + (-2+3)k = -2i - 4j + k$.
The work done $W$ is given by the dot product of force $\vec{F}$ and displacement $\vec{d}$:
$W = \vec{F} \cdot \vec{d} = (i - 3j + 5k) \cdot (-2i - 4j + k)$.
$W = (1 \times -2) + (-3 \times -4) + (5 \times 1) = -2 + 12 + 5 = 15 \text{ units}$.

(A) The work done $W$ is given by the dot product of the force vector $\overrightarrow{F}$ and the displacement vector $\overrightarrow{d}$.
First,calculate the displacement vector $\overrightarrow{d} = \vec{r_2} - \vec{r_1}$.
$\overrightarrow{d} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - \hat{k}) = (2-1)\hat{i} + (-1-1)\hat{j} + (1-(-1))\hat{k} = \hat{i} - 2\hat{j} + 2\hat{k}$.
Now,calculate the work done $W = \overrightarrow{F} \cdot \overrightarrow{d}$.
$W = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (\hat{i} - 2\hat{j} + 2\hat{k})$.
$W = (1)(1) + (2)(-2) + (3)(2) = 1 - 4 + 6 = 3$.
Thus,the work done is $3$ units.

(A) The force vector is given by $F = 2i - 3j + 2k$.
The initial position vector is $r_1 = 3i + 4j + 5k$ and the final position vector is $r_2 = 1i + 2j + 3k$.
The displacement vector $d$ is given by $d = r_2 - r_1 = (1 - 3)i + (2 - 4)j + (3 - 5)k = -2i - 2j - 2k$.
The work done $W$ is the dot product of force and displacement: $W = F \cdot d$.
$W = (2i - 3j + 2k) \cdot (-2i - 2j - 2k) = (2 \times -2) + (-3 \times -2) + (2 \times -2) = -4 + 6 - 4 = -2$.
The magnitude of the work done is $|W| = |-2| = 2$ $unit$.

(C) The resultant force $\vec{F}$ is the sum of the individual forces:
$\vec{F} = (3i + 2j + 5k) + (2i + j - 3k) = (3+2)i + (2+1)j + (5-3)k = 5i + 3j + 2k$.
The displacement vector $\vec{d}$ is the difference between the final position vector and the initial position vector:
$\vec{d} = (4i - 3j + 7k) - (2i - j - 3k) = (4-2)i + (-3 - (-1))j + (7 - (-3))k = 2i - 2j + 10k$.
Work done $W$ is the dot product of the force and displacement vectors:
$W = \vec{F} \cdot \vec{d} = (5i + 3j + 2k) \cdot (2i - 2j + 10k)$
$W = (5 \times 2) + (3 \times -2) + (2 \times 10)$
$W = 10 - 6 + 20 = 24 \, \text{units}$.

(D) The resultant force $\vec{F}$ is the sum of the two forces:
$\vec{F} = (3i + 2j - 3k) + (2i + 4j + 2k) = 5i + 6j - k$.
The displacement vector $\vec{d}$ is the difference between the final and initial position vectors:
$\vec{d} = (5i + 4j + 2k) - (i + 2j + k) = 4i + 2j + k$.
The work done $W$ is the dot product of the force and displacement vectors:
$W = \vec{F} \cdot \vec{d} = (5i + 6j - k) \cdot (4i + 2j + k)$.
$W = (5 \times 4) + (6 \times 2) + (-1 \times 1) = 20 + 12 - 1 = 31 \text{ units}$.

(C) The work done $W$ is defined as the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
Given:
$\vec{F} = 2i - j - k$
$\vec{d} = 3i + 2j - 5k$
$W = \vec{F} \cdot \vec{d}$
$W = (2i - j - k) \cdot (3i + 2j - 5k)$
$W = (2 \times 3) + (-1 \times 2) + (-1 \times -5)$
$W = 6 - 2 + 5$
$W = 9 \text{ units}$.

(B) The work done $W$ is given by the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
First,find the unit vector in the direction of $2i - 2j + k$:
$\hat{u} = \frac{2i - 2j + k}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{2i - 2j + k}{\sqrt{4 + 4 + 1}} = \frac{2i - 2j + k}{3}$.
The force vector $\vec{F}$ is the magnitude multiplied by the unit vector:
$\vec{F} = 5 \times \left( \frac{2i - 2j + k}{3} \right) = \frac{5}{3}(2i - 2j + k)$.
The displacement vector $\vec{d}$ is the difference between the final and initial positions:
$\vec{d} = (5i + 3j + 7k) - (i + 2j + 3k) = (5-1)i + (3-2)j + (7-3)k = 4i + j + 4k$.
The work done $W = \vec{F} \cdot \vec{d}$:
$W = \frac{5}{3}(2i - 2j + k) \cdot (4i + j + 4k)$
$W = \frac{5}{3} \times [(2 \times 4) + (-2 \times 1) + (1 \times 4)]$
$W = \frac{5}{3} \times [8 - 2 + 4] = \frac{5}{3} \times 10 = \frac{50}{3}$ units.

(C) The resultant force $\overrightarrow{F}$ is the sum of the individual forces:
$\overrightarrow{F} = (4i + j - 3k) + (3i + j - k) = (4+3)i + (1+1)j + (-3-1)k = 7i + 2j - 4k$.
The displacement vector $\overrightarrow{d}$ is the difference between the final position vector and the initial position vector:
$\overrightarrow{d} = (5i + 4j + k) - (i + 2j + 3k) = (5-1)i + (4-2)j + (1-3)k = 4i + 2j - 2k$.
The work done $W$ is the dot product of the force and displacement vectors:
$W = \overrightarrow{F} \cdot \overrightarrow{d} = (7i + 2j - 4k) \cdot (4i + 2j - 2k)$.
$W = (7 \times 4) + (2 \times 2) + (-4 \times -2) = 28 + 4 + 8 = 40 \text{ units}$.

(A) The scalar projection of a vector $\vec{a}$ on a vector $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Let $\vec{a} = xi - j + k$ and $\vec{b} = 2i - j + 5k$.
The dot product $\vec{a} \cdot \vec{b} = (x)(2) + (-1)(-1) + (1)(5) = 2x + 1 + 5 = 2x + 6$.
The magnitude of vector $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 5^2} = \sqrt{4 + 1 + 25} = \sqrt{30}$.
Given that the scalar projection is $\frac{1}{\sqrt{30}}$, we have $\frac{2x + 6}{\sqrt{30}} = \frac{1}{\sqrt{30}}$.
Equating the numerators, we get $2x + 6 = 1$.
$2x = 1 - 6 = -5$.
Therefore, $x = \frac{-5}{2}$.