Basic , Modulus and Algebra of vectors Questions in English

(A) Given the vector equation $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$.
Substituting the given vectors,we have $\alpha(1, 2, 3) + \beta(2, 3, 1) + \gamma(3, 1, 2) = -3(1, 0, -1)$.
This results in the system of linear equations:
$1) \alpha + 2\beta + 3\gamma = -3$
$2) 2\alpha + 3\beta + \gamma = 0$
$3) 3\alpha + \beta + 2\gamma = 3$
Adding all three equations: $(\alpha + 2\alpha + 3\alpha) + (2\beta + 3\beta + \beta) + (3\gamma + \gamma + 2\gamma) = -3 + 0 + 3 \Rightarrow 6\alpha + 6\beta + 6\gamma = 0 \Rightarrow \alpha + \beta + \gamma = 0$.
From equation $(2)$,$\gamma = -2\alpha - 3\beta$. Substituting this into $\alpha + \beta + \gamma = 0$ gives $\alpha + \beta - 2\alpha - 3\beta = 0 \Rightarrow -\alpha - 2\beta = 0 \Rightarrow \alpha = -2\beta$.
Substituting $\alpha = -2\beta$ into equation $(1)$: $(-2\beta) + 2\beta + 3\gamma = -3 \Rightarrow 3\gamma = -3 \Rightarrow \gamma = -1$.
Since $\alpha + \beta + \gamma = 0$,we have $-2\beta + \beta - 1 = 0 \Rightarrow -\beta = 1 \Rightarrow \beta = -1$.
Then $\alpha = -2(-1) = 2$.
Thus,the triplet $(\alpha, \beta, \gamma)$ is $(2, -1, -1)$.

(B) Given that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$.
Squaring both sides,we get $|\vec{u} + \vec{v} + \vec{w}|^2 = |\vec{0}|^2$.
Using the identity $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$,we have:
$|\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = 0$.
Substituting the given magnitudes $|\vec{u}| = 3$,$|\vec{v}| = 4$,and $|\vec{w}| = 5$:
$3^2 + 4^2 + 5^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = 0$.
$9 + 16 + 25 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = 0$.
$50 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = 0$.
$2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = -50$.
Therefore,$\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} = -25$.

(A) Since $\vec{a}$ and $\vec{b}$ are linearly dependent,we can write $\vec{b} = \lambda \vec{a}$ for some scalar $\lambda \neq 0$.
Substituting this into the given equation:
$\frac{|\vec{a} + \lambda \vec{a}|}{|\vec{a} - \lambda \vec{a}|} = 2$
$\frac{|1 + \lambda| |\vec{a}|}{|1 - \lambda| |\vec{a}|} = 2$
Since $\vec{a} \neq 0$,we have $\frac{|1 + \lambda|}{|1 - \lambda|} = 2$.
Squaring both sides:
$\frac{(1 + \lambda)^2}{(1 - \lambda)^2} = 4$
$(1 + \lambda)^2 = 4(1 - \lambda)^2$
$1 + 2\lambda + \lambda^2 = 4(1 - 2\lambda + \lambda^2)$
$1 + 2\lambda + \lambda^2 = 4 - 8\lambda + 4\lambda^2$
$3\lambda^2 - 10\lambda + 3 = 0$
$(3\lambda - 1)(\lambda - 3) = 0$
Thus,$\lambda = 3$ or $\lambda = \frac{1}{3}$.
Given the condition $|\vec{b}| > |\vec{a}|$,we have $|\lambda \vec{a}| > |\vec{a}|$,which implies $|\lambda| > 1$.
Therefore,$\lambda = 3$ is the only valid solution.
Hence,$\vec{b} = 3\vec{a}$.

(A) Given the equation $\vec{r} \times \vec{b} = \vec{r} \times \vec{c}$.
This can be rewritten as $\vec{r} \times \vec{b} - \vec{r} \times \vec{c} = \vec{0}$,which implies $\vec{r} \times (\vec{b} - \vec{c}) = \vec{0}$.
Calculate $\vec{b} - \vec{c} = (\hat{i} + 2\hat{j} + \hat{k}) - (3\hat{i} + 2\hat{k}) = -2\hat{i} + 2\hat{j} - \hat{k}$.
Since $\vec{r} \times (\vec{b} - \vec{c}) = \vec{0}$,the vector $\vec{r}$ must be parallel to $(\vec{b} - \vec{c})$.
Thus,$\vec{r} = \lambda (-2\hat{i} + 2\hat{j} - \hat{k})$.
For $\vec{r}$ to be a unit vector,we have $|\vec{r}| = 1$,so $|\lambda| \sqrt{(-2)^2 + 2^2 + (-1)^2} = 1$.
$|\lambda| \sqrt{4 + 4 + 1} = 1 \Rightarrow |\lambda| \sqrt{9} = 1 \Rightarrow 3|\lambda| = 1 \Rightarrow \lambda = \pm \frac{1}{3}$.
Therefore,$\hat{r} = \pm \frac{1}{3} (-2\hat{i} + 2\hat{j} - \hat{k}) = \pm \left( \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} \right)$.

(B) Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two unit vectors,so $|\overrightarrow{a}| = |\overrightarrow{b}| = 1$. Let $\theta$ be the angle between them.
The magnitude of their sum is $|\overrightarrow{a} + \overrightarrow{b}| = \sqrt{1^2 + 1^2 + 2(1)(1)\cos \theta} = \sqrt{2 + 2\cos \theta} = 2\cos(\theta/2)$.
The magnitude of their difference is $|\overrightarrow{a} - \overrightarrow{b}| = \sqrt{1^2 + 1^2 - 2(1)(1)\cos \theta} = \sqrt{2 - 2\cos \theta} = 2\sin(\theta/2)$.
Given the condition: $|\overrightarrow{a} - \overrightarrow{b}| < |\overrightarrow{a} + \overrightarrow{b}| < \sqrt{3}|\overrightarrow{a} - \overrightarrow{b}|$.
First part: $|\overrightarrow{a} - \overrightarrow{b}| < |\overrightarrow{a} + \overrightarrow{b}| \Rightarrow 2\sin(\theta/2) < 2\cos(\theta/2) \Rightarrow \tan(\theta/2) < 1 \Rightarrow \theta/2 < \pi/4 \Rightarrow \theta < \pi/2$.
Second part: $|\overrightarrow{a} + \overrightarrow{b}| < \sqrt{3}|\overrightarrow{a} - \overrightarrow{b}| \Rightarrow 2\cos(\theta/2) < \sqrt{3} \cdot 2\sin(\theta/2) \Rightarrow \cot(\theta/2) < \sqrt{3} \Rightarrow \tan(\theta/2) > 1/\sqrt{3} \Rightarrow \theta/2 > \pi/6 \Rightarrow \theta > \pi/3$.
Combining these,we get $\theta \in \left(\frac{\pi}{3}, \frac{\pi}{2}\right)$.

(C) By the triangle law of vector addition,we have:
$\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = \vec{AE}$ ........$(1)$
Similarly,for the other path:
$\vec{AH} + \vec{HG} + \vec{GF} + \vec{FE} = \vec{AE}$ ........$(2)$
Adding Equation $(1)$ and Equation $(2)$:
$\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{AH} + \vec{HG} + \vec{GF} + \vec{FE} = \vec{AE} + \vec{AE} = 2\vec{AE}$

(D) The projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by $\frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|}$.
The projection of vector $\overrightarrow{b}$ on $\overrightarrow{a}$ is given by $\frac{\overrightarrow{b} \cdot \overrightarrow{a}}{|\overrightarrow{a}|}$.
The ratio of the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ to the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is:
$\frac{\frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|}}{\frac{\overrightarrow{b} \cdot \overrightarrow{a}}{|\overrightarrow{a}|}} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} \times \frac{|\overrightarrow{a}|}{\overrightarrow{b} \cdot \overrightarrow{a}}$
Since the dot product is commutative,$\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{a}$.
Therefore,the ratio simplifies to $\frac{|\overrightarrow{a}|}{|\overrightarrow{b}|} = \frac{2}{3}$.

(D) The position vectors of points $A, B, C$ are $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{i} - \hat{j}$,and $\vec{c} = p\hat{i} - q\hat{j} + r\hat{k}$.
For points $A, B, C$ to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be parallel,i.e.,$\overrightarrow{AC} = \lambda \overrightarrow{AB}$ for some scalar $\lambda$.
Calculate $\overrightarrow{AB} = \vec{b} - \vec{a} = (\hat{i} - \hat{j}) - (\hat{i} + \hat{j}) = -2\hat{j}$.
Calculate $\overrightarrow{AC} = \vec{c} - \vec{a} = (p-1)\hat{i} - (q+1)\hat{j} + r\hat{k}$.
Since $\overrightarrow{AC} = \lambda \overrightarrow{AB}$,we have $(p-1)\hat{i} - (q+1)\hat{j} + r\hat{k} = \lambda(-2\hat{j}) = -2\lambda\hat{j}$.
Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$:
$1$) $p-1 = 0 \implies p = 1$.
$2$) $-(q+1) = -2\lambda \implies q+1 = 2\lambda$.
$3$) $r = 0$.
From option $D$,if $p=1, q=2, r=0$,then $2+1 = 2\lambda \implies 3 = 2\lambda \implies \lambda = 1.5$. This is a valid scalar,so the points are collinear. Thus,$p=1, q=2, r=0$ is a correct possibility.

(B) Given $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$.
Since $3p = 2q$,we have:
$3[(x + 4y)\vec{a} + (2x + y + 1)\vec{b}] = 2[(y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}]$
$(3x + 12y)\vec{a} + (6x + 3y + 3)\vec{b} = (2y - 4x + 4)\vec{a} + (4x - 6y - 2)\vec{b}$
Since $\vec{a}$ and $\vec{b}$ are non-collinear,we can equate the coefficients:
$1) 3x + 12y = 2y - 4x + 4 \Rightarrow 7x + 10y = 4$
$2) 6x + 3y + 3 = 4x - 6y - 2 \Rightarrow 2x + 9y = -5$
Multiplying equation $(1)$ by $2$ and equation $(2)$ by $7$:
$14x + 20y = 8$
$14x + 63y = -35$
Subtracting the equations: $(63y - 20y) = -35 - 8 \Rightarrow 43y = -43 \Rightarrow y = -1$
Substituting $y = -1$ into $2x + 9y = -5$:
$2x + 9(-1) = -5 \Rightarrow 2x - 9 = -5 \Rightarrow 2x = 4 \Rightarrow x = 2$
Thus,$x = 2$ and $y = -1$.

(B) Given that $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $|\vec{a} - \vec{b}| = 1$.
We know the identity $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Since $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,we have:
$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}| |\vec{b}| \cos \theta$.
Substituting the given values:
$1^2 = 1^2 + 1^2 - 2(1)(1) \cos \theta$
$1 = 1 + 1 - 2 \cos \theta$
$1 = 2 - 2 \cos \theta$
$2 \cos \theta = 1$
$\cos \theta = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,the angle $\theta = \frac{\pi}{3}$.

(C) Let $\vec{G}$ be the position vector of the centroid of $\triangle ABC$. We know that $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.
Let $\vec{H}$ be the position vector of the orthocentre and $\vec{P}$ be the position vector of the circumcentre. We are given $\vec{P} = \frac{\vec{a} + \vec{b} + \vec{c}}{4}$.
In any triangle,the centroid $\vec{G}$ divides the line segment joining the orthocentre $\vec{H}$ and the circumcentre $\vec{P}$ in the ratio $2:1$. That is,$\vec{G} = \frac{2\vec{P} + 1\vec{H}}{2+1} = \frac{2\vec{P} + \vec{H}}{3}$.
Therefore,$3\vec{G} = 2\vec{P} + \vec{H}$,which implies $\vec{H} = 3\vec{G} - 2\vec{P}$.
Substituting the values of $\vec{G}$ and $\vec{P}$:
$\vec{H} = 3\left(\frac{\vec{a} + \vec{b} + \vec{c}}{3}\right) - 2\left(\frac{\vec{a} + \vec{b} + \vec{c}}{4}\right)$
$\vec{H} = (\vec{a} + \vec{b} + \vec{c}) - \frac{\vec{a} + \vec{b} + \vec{c}}{2}$
$\vec{H} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$.

(C) Given $|2\vec{a} - \vec{b}| = 5$.
Squaring both sides,we get $|2\vec{a} - \vec{b}|^2 = 25$.
Using the property $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we have:
$4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b}) = 25$.
Substituting $|\vec{a}| = 2$ and $|\vec{b}| = 3$:
$4(2)^2 + (3)^2 - 4(\vec{a} \cdot \vec{b}) = 25$.
$16 + 9 - 4(\vec{a} \cdot \vec{b}) = 25$.
$25 - 4(\vec{a} \cdot \vec{b}) = 25 \implies 4(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0$.
Now,we need to find $|2\vec{a} + \vec{b}|$.
$|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b})$.
Substituting the values:
$|2\vec{a} + \vec{b}|^2 = 4(4) + 9 + 4(0) = 16 + 9 = 25$.
Therefore,$|2\vec{a} + \vec{b}| = \sqrt{25} = 5$.

(D) Let $\vec{c} = a\hat{i} + b\hat{j} + c\hat{k}$.
Given $\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$,this implies that $\vec{c}$ is parallel to the vector $\vec{v} = \hat{i} + 2\hat{j} + 5\hat{k}$.
Thus,$\vec{c} = k(\hat{i} + 2\hat{j} + 5\hat{k})$ for some scalar $k$.
Given $|\vec{c}|^2 = 60$,we have $k^2(1^2 + 2^2 + 5^2) = 60$.
$k^2(1 + 4 + 25) = 60 \Rightarrow 30k^2 = 60 \Rightarrow k^2 = 2 \Rightarrow k = \pm\sqrt{2}$.
Taking $k = \sqrt{2}$,we get $\vec{c} = \sqrt{2}\hat{i} + 2\sqrt{2}\hat{j} + 5\sqrt{2}\hat{k}$.
Now,calculate the dot product $\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$:
$= (\sqrt{2}\hat{i} + 2\sqrt{2}\hat{j} + 5\sqrt{2}\hat{k}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$
$= \sqrt{2}(-7) + 2\sqrt{2}(2) + 5\sqrt{2}(3)$
$= -7\sqrt{2} + 4\sqrt{2} + 15\sqrt{2} = 12\sqrt{2}$.

(B) Given that $\hat{x}, \hat{y}, \hat{z}$ are unit vectors,so $|\hat{x}| = |\hat{y}| = |\hat{z}| = 1$.
We know that for any vectors,$|\hat{x} + \hat{y} + \hat{z}|^2 \geq 0$.
Expanding this,we get $|\hat{x}|^2 + |\hat{y}|^2 + |\hat{z}|^2 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq 0$.
Substituting the magnitudes,$1 + 1 + 1 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq 0$,which implies $3 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq 0$.
Thus,$2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq -3$.
Now,consider the expression $S = |\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$.
$S = (\hat{x} \cdot \hat{x} + \hat{y} \cdot \hat{y} + 2\hat{x} \cdot \hat{y}) + (\hat{y} \cdot \hat{y} + \hat{z} \cdot \hat{z} + 2\hat{y} \cdot \hat{z}) + (\hat{z} \cdot \hat{z} + \hat{x} \cdot \hat{x} + 2\hat{z} \cdot \hat{x})$.
$S = 2(|\hat{x}|^2 + |\hat{y}|^2 + |\hat{z}|^2) + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x})$.
$S = 2(1 + 1 + 1) + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) = 6 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x})$.
Since $2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq -3$,the minimum value of $S$ is $6 - 3 = 3$.

(B) Two vectors $\vec{u}$ and $\vec{v}$ are collinear if there exists a non-zero scalar $k$ such that $\vec{u} = k\vec{v}$.
Given $\vec{u} = (\alpha - 2)\vec{a} + \vec{b}$ and $\vec{v} = (2 + 3\alpha)\vec{a} - 3\vec{b}$.
Since $\vec{u}$ and $\vec{v}$ are collinear,we have $(\alpha - 2)\vec{a} + \vec{b} = k((2 + 3\alpha)\vec{a} - 3\vec{b})$.
Rearranging the terms,we get $(\alpha - 2 - k(2 + 3\alpha))\vec{a} + (1 + 3k)\vec{b} = 0$.
Since $\vec{a}$ and $\vec{b}$ are non-collinear,their coefficients must be zero.
Thus,$1 + 3k = 0 \Rightarrow k = -\frac{1}{3}$.
Substituting $k = -\frac{1}{3}$ into the coefficient of $\vec{a}$:
$\alpha - 2 - (-\frac{1}{3})(2 + 3\alpha) = 0$.
$\alpha - 2 + \frac{2}{3} + \alpha = 0$.
$2\alpha = 2 - \frac{2}{3} = \frac{4}{3}$.
$\alpha = \frac{2}{3}$.

(A) Given vectors are $\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$.
Since $\vec{a}$ and $\vec{b}$ are non-collinear,$\vec{\alpha}$ and $\vec{\beta}$ are collinear if and only if the ratio of the coefficients of $\vec{a}$ and $\vec{b}$ are equal.
This implies $\frac{\lambda - 2}{4\lambda - 2} = \frac{1}{3}$.
Cross-multiplying,we get $3(\lambda - 2) = 1(4\lambda - 2)$.
$3\lambda - 6 = 4\lambda - 2$.
Rearranging the terms,$3\lambda - 4\lambda = -2 + 6$.
$-\lambda = 4$.
Therefore,$\lambda = -4$.

(N/A) To represent the displacement of $40 \, km,$ $30^\circ$ west of south graphically:
$1$. Draw a coordinate system with North $(N)$,South $(S)$,East $(E)$,and West $(W)$ directions.
$2$. Choose a suitable scale,for example,$1 \, cm = 10 \, km$. Thus,$40 \, km$ corresponds to a length of $4 \, cm$.
$3$. Start from the origin $O$ and draw a line segment $OP$ of length $4 \, cm$ in the direction $30^\circ$ west of south (i.e.,starting from the South axis and moving $30^\circ$ towards the West).
$4$. The vector $\vec{OP}$ represents the required displacement of $40 \, km,$ $30^\circ$ west of south.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
Time is a physical quantity that is measured in seconds. It does not have a specific direction associated with it.
Therefore,$5 \text{ seconds}$ is a scalar quantity.

(A) The given measure is $1000 \, cm^{3}$,which represents volume.
Volume is a physical quantity that has only magnitude and no direction.
Therefore,volume is a scalar quantity.

(B) The measure $10 \text{ Newton}$ represents force.
Force is a physical quantity that has both magnitude and direction.
Therefore,force is a vector quantity.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
$A$ vector quantity is defined as a physical quantity that has both magnitude and direction.
The given measure is $30 \, km/hr$,which represents speed.
Speed is the magnitude of velocity and does not have a specified direction.
Therefore,$30 \, km/hr$ is a scalar quantity.

(A) Density is defined as mass per unit volume. Since mass and volume are both scalar quantities,their ratio,density,is also a scalar quantity. It has magnitude but no specific direction associated with it. Therefore,$10 \, g/cm^3$ is a scalar quantity.

(B) scalar quantity is defined only by its magnitude,whereas a vector quantity is defined by both magnitude and direction.
The given measure is $20 \text{ m/s}$ towards north.
Since it has both magnitude $(20 \text{ m/s})$ and direction (towards north),it is a vector quantity.
Therefore,the measure is a vector.

(N/A) Two or more vectors are said to be collinear if they are parallel to the same line,irrespective of their magnitudes and directions.
Looking at the figure,the vectors $\vec{c}$ and $\vec{d}$ lie along the same line (or are parallel to the same line).
Therefore,the collinear vectors are $\vec{c}$ and $\vec{d}$.

(N/A) Two vectors are said to be equal if they have the same magnitude and the same direction.
$1$. Vector $\vec{a}$ has a magnitude of $1 \text{ unit}$ and points in a specific direction.
$2$. Vector $\vec{b}$ has a magnitude of $2 \text{ units}$ and points in a different direction.
$3$. Vector $\vec{c}$ has a magnitude of $1 \text{ unit}$ and points in the same direction as $\vec{a}$.
$4$. Vector $\vec{d}$ has a magnitude of $2 \text{ units}$ and points in the opposite direction to $\vec{b}$.
Comparing these,vectors $\vec{a}$ and $\vec{c}$ have the same magnitude $(1 \text{ unit})$ and the same direction.
Therefore,the equal vectors are $\vec{a}$ and $\vec{c}$.

(N/A) Coinitial vectors are those vectors which have the same initial point.
Looking at the figure,we observe that vectors $\vec{b}$,$\vec{c}$,and $\vec{d}$ all originate from the same point.
Therefore,the coinitial vectors are $\vec{b}$,$\vec{c}$,and $\vec{d}$.

(N/A) To represent the displacement of $40 \, km$,$30^{\circ}$ east of north graphically:
$1$. Draw a coordinate system with the North,South,East,and West directions.
$2$. Let the origin $O$ represent the starting point.
$3$. Draw a ray $OP$ starting from $O$ such that it makes an angle of $30^{\circ}$ with the North direction towards the East.
$4$. Using a scale of $1 \, cm = 10 \, km$,mark the length of the vector $OP$ as $4 \, cm$ to represent $40 \, km$.
$5$. The vector $\overrightarrow{OP}$ represents the required displacement.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
$10 \text{ kg}$ represents mass,which is a measure of the amount of matter in an object.
Since mass has only magnitude and does not require a direction to be specified,it is a scalar quantity.
Therefore,$10 \text{ kg}$ is a scalar.

(B) scalar quantity is defined only by its magnitude,whereas a vector quantity is defined by both magnitude and direction.
In the given measure,$2 \text{ meters}$ represents the magnitude and $\text{north-west}$ represents the direction.
Since the measure involves both magnitude and direction,it is a vector quantity.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
$A$ vector quantity is defined as a physical quantity that has both magnitude and direction.
The measure $40^{o}$ represents an angle,which has only magnitude and no specific direction associated with it.
Therefore,$40^{o}$ is a scalar quantity.

(A) $40$ $watt$ is a scalar quantity because it has only magnitude and no direction associated with it.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
Electric charge,measured in $Coulomb$,is a scalar quantity because it does not have a specific direction associated with it.
Therefore,$10^{-19} \text{ Coulomb}$ is a scalar quantity.

(B) $20 \, m/s^{2}$ represents acceleration.
Acceleration is a vector quantity because it possesses both magnitude and direction.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
$A$ vector quantity is defined as a physical quantity that has both magnitude and direction.
Time period represents the duration of an event,which is described solely by its magnitude (e.g.,$5 \text{ seconds}$).
Since it does not possess any specific direction,time period is a scalar quantity.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
Distance represents the total path length covered by an object between two points.
Since distance does not have a specific direction associated with it,it is classified as a scalar quantity.

(B) scalar quantity is defined only by its magnitude,whereas a vector quantity is defined by both its magnitude and direction.
Force is a physical quantity that requires both magnitude (strength of the push or pull) and a specific direction to be fully described.
Therefore,force is a vector quantity.

(B) scalar quantity is defined only by its magnitude,whereas a vector quantity is defined by both its magnitude and direction.
Velocity is defined as the rate of change of displacement,which inherently includes direction.
Therefore,velocity is a vector quantity.

(N/A) Co-initial vectors are vectors that originate from the same initial point.
Looking at the square,vectors $\vec{a}$ and $\vec{d}$ originate from the top-left corner of the square.
Therefore,$\vec{a}$ and $\vec{d}$ are co-initial vectors.

(N/A) Two vectors are said to be equal if they have the same magnitude and the same direction.
In the given square,vectors $\overrightarrow{a}$ and $\overrightarrow{c}$ have the same magnitude (side length of the square) but opposite directions.
Vectors $\overrightarrow{d}$ and $\overrightarrow{b}$ have the same magnitude (side length of the square) but opposite directions.
Looking at the directions of the vectors:
- $\overrightarrow{a}$ points to the right.
- $\overrightarrow{c}$ points to the left.
- $\overrightarrow{d}$ points downwards.
- $\overrightarrow{b}$ points downwards.
Therefore,vectors $\overrightarrow{d}$ and $\overrightarrow{b}$ are equal because they have the same magnitude and the same direction.

(N/A) Two vectors are said to be collinear if they are parallel to the same line,irrespective of their magnitudes and directions.
In the given square,vectors $\vec{a}$ and $\vec{c}$ are parallel to each other but have opposite directions. Therefore,they are collinear.
However,they are not equal because equal vectors must have the same magnitude and the same direction.
Thus,the vectors that are collinear but not equal are $\vec{a}$ and $\vec{c}$.

(A) True
Two vectors are said to be collinear if they are parallel to the same line,irrespective of their magnitudes and directions.
Since $-\vec{a}$ is a scalar multiple of $\vec{a}$ (specifically,$-\vec{a} = -1 \times \vec{a}$),the vectors $\vec{a}$ and $-\vec{a}$ are parallel to each other.
Therefore,$\vec{a}$ and $-\vec{a}$ are collinear.

(B) The statement is $False$.
Collinear vectors are defined as vectors that are parallel to the same line,irrespective of their magnitudes.
For example,if $\vec{a} = 2\hat{i}$ and $\vec{b} = 3\hat{i}$,both are collinear because they are parallel to the $x$-axis,but their magnitudes are $|\vec{a}| = 2$ and $|\vec{b}| = 3$,which are not equal.

(B) The statement is False.
Two vectors are said to be collinear if they are parallel to the same line,irrespective of their magnitudes.
Having the same magnitude does not imply that the vectors are parallel.
For example,consider vector $\vec{a} = \hat{i}$ and vector $\vec{b} = \hat{j}$.
Both vectors have a magnitude of $1$,but they are not collinear because they are not parallel to the same line.

(B) The statement is False.
Two vectors are said to be equal if they have the same magnitude and the same direction.
Collinear vectors are vectors that are parallel to the same line,irrespective of their magnitudes and directions.
Even if two collinear vectors have the same magnitude,they may have opposite directions (e.g.,$\vec{a}$ and $-\vec{a}$),which makes them unequal.

(A) Two vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ are equal if and only if their corresponding components are equal,i.e.,$a_1 = b_1, a_2 = b_2$ and $a_3 = b_3$.
Given vectors are $\vec{a} = x \hat{i} + 2 \hat{j} + z \hat{k}$ and $\vec{b} = 2 \hat{i} + y \hat{j} + \hat{k}$.
Comparing the components of $\vec{a}$ and $\vec{b}$:
For the $\hat{i}$ component: $x = 2$.
For the $\hat{j}$ component: $2 = y$,which means $y = 2$.
For the $\hat{k}$ component: $z = 1$.
Therefore,the values are $x = 2, y = 2, z = 1$.

(N/A) Given vectors are $\vec{a} = \hat{i} + 2\hat{j}$ and $\vec{b} = 2\hat{i} + \hat{j}$.
The magnitude of vector $\vec{a}$ is $|\vec{a}| = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5}$.
The magnitude of vector $\vec{b}$ is $|\vec{b}| = \sqrt{2^{2} + 1^{2}} = \sqrt{4 + 1} = \sqrt{5}$.
Since $|\vec{a}| = \sqrt{5}$ and $|\vec{b}| = \sqrt{5}$,we have $|\vec{a}| = |\vec{b}|$.
Two vectors are equal if and only if their corresponding components are equal. Here,the components of $\vec{a}$ are $(1, 2)$ and the components of $\vec{b}$ are $(2, 1)$.
Since $(1, 2) \neq (2, 1)$,the vectors $\vec{a}$ and $\vec{b}$ are not equal.

(A) The unit vector in the direction of a vector $\vec{a}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
First,calculate the magnitude of vector $\vec{a}$:
$|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
Now,divide the vector $\vec{a}$ by its magnitude:
$\hat{a} = \frac{1}{\sqrt{14}}(2\hat{i} + 3\hat{j} + \hat{k}) = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}$.

(A) The magnitude of the vector $\vec{a} = \hat{i} - 2\hat{j}$ is $|\vec{a}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
The unit vector in the direction of $\vec{a}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{5}}(\hat{i} - 2\hat{j}) = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}$.
$A$ vector of magnitude $7$ in the direction of $\vec{a}$ is $7\hat{a} = 7 \left( \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j} \right) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}$.

(A) Let $\vec{c} = \vec{a} + \vec{b}$.
Given $\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.
Then $\vec{c} = (2+2)\hat{i} + (2+1)\hat{j} + (-5+3)\hat{k} = 4\hat{i} + 3\hat{j} - 2\hat{k}$.
The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.
The unit vector in the direction of $\vec{c}$ is given by $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.
$\hat{c} = \frac{1}{\sqrt{29}}(4\hat{i} + 3\hat{j} - 2\hat{k}) = \frac{4}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{2}{\sqrt{29}}\hat{k}$.

(A) Since the vector is directed from $P$ to $Q$,$P$ is the initial point and $Q$ is the terminal point.
The vector $\overrightarrow{PQ}$ is given by the difference of the position vectors of the terminal point and the initial point:
$\overrightarrow{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$
Given $P(2, 3, 0)$ and $Q(-1, -2, -4)$:
$\overrightarrow{PQ} = (-1 - 2)\hat{i} + (-2 - 3)\hat{j} + (-4 - 0)\hat{k}$
$\overrightarrow{PQ} = -3\hat{i} - 5\hat{j} - 4\hat{k}$