The vector having initial and terminal points as $(2, 5, 0)$ and $(-3, 7, 4)$, respectively is (A) $-\hat{i}+12\hat{j}+4\hat{k}$ (B) $5\hat{i}+2\hat{j}-4\hat{k}$ (C) $-5\hat{i}+2\hat{j}+4\hat{k}$ (D) $\hat{i}+\hat{j}+\hat{k}$

The vector having initial and terminal points as (2, 5, 0) and (-3, 7, 4), respectively is (A) -hati+12hatj+4h

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(2, 5, 0)

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The vector from point $A(2,5,0)$ to point $B(-3,7,4)$ is found by subtracting coordinates: terminal minus initial. The result is $-5\hat{i}+2\hat{j}+4\hat{k}$ , which matches option (C).

Why Direction Vectors Work This Way

A vector is defined by its displacement — how much you move along each axis to go from the start to the end. If you think of walking from point $A$ to point $B$ , the vector $\overrightarrow{AB}$ tells you: "go left/right this much, forward/backward this much, up/down this much."

The neat trick is that you don't need geometry or a diagram. Just subtract the coordinates of the initial point from the coordinates of the terminal point, component by component. That subtraction gives the net change in each direction.

A(x_1, y_1, z_1)

Step-by-Step

Identify the points

Initial point: $A = (2, 5, 0)$

Terminal point: $B = (-3, 7, 4)$
Subtract coordinates: terminal minus initial
- $x$ -component: $-3 - 2 = -5$
- $y$ -component: $7 - 5 = 2$
- $z$ -component: $4 - 0 = 4$
Write the vector in component form

$\overrightarrow{AB} = -5\hat{i} + 2\hat{j} + 4\hat{k}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 65/2/11 markMCQ
$x$
›Reveal solutionSolution
The vector makes equal angles with the $x$ and $y$ axes and is perpendicular to the $z$ axis, so its $z$ -component is zero and its $x$ and $y$ components are equal. With magnitude 3, each component is $\frac{3}{\sqrt{2}}$ , giving the vector $\frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}$ , which matches option (C) after rationalising.

The key idea here is that a vector’s direction cosines tell you how it’s oriented relative to the axes. If a vector makes equal angles with the $x$ and $y$ axes, its direction cosines (and hence its components) along those axes are equal. And if it’s perpendicular to the $z$ axis, its $z$ -component is zero — the vector lies entirely in the $xy$ -plane.

So we’re looking for a vector of the form $a\hat{i} + a\hat{j} + 0\hat{k}$ , where $a$ is the (equal) $x$ and $y$ components. The magnitude condition then fixes $a$ .

Let’s work through it.

Set up the vector form.

Let the vector be $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ .

Perpendicular to the $z$ axis means $v_z = 0$ .

Equal angles with $x$ and $y$ axes means the direction cosines $\cos\alpha$ and $\cos\beta$ are equal. Since direction cosines are proportional to the components, we have $v_x = v_y$ .

So $\vec{v} = a\hat{i} + a\hat{j}$ , where $a = v_x = v_y$ .

Apply the magnitude condition.

The magnitude is given as 3:

$|\vec{v}| = \sqrt{a^2 + a^2 + 0^2} = \sqrt{2a^2} = |a|\sqrt{2} = 3.$
Since magnitude is positive, $a$ must be positive (the vector’s direction is fixed by the signs, but the problem doesn’t specify a sign, so we take the positive one).
$a\sqrt{2} = 3 \quad\Rightarrow\quad a = \frac{3}{\sqrt{2}}.$

Write the vector.

$\vec{v} = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}.$
This is not yet in the form of any option — but option (C) is $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$ . Notice that $\frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ (multiply numerator and denominator by $\sqrt{2}$ ). So they are identical. …
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CBSE 2026Set V11 markMCQ
$x$
(a) $0, k, 0$
(b) $0, 0, k$
(c) $k, 0, 0$
(d) $k, k, k$
›Reveal solutionSolution

The $x$ -axis points along $\hat i=(1,0,0)$ , so its direction ratios are $k,0,0$ ; answer (c).

The positive $x$ -axis is directed along the unit vector $\hat i=(1,0,0)$ . Direction ratios are any nonzero scalar multiple of the d …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Direction ratios of the straight line vector r = 2i - 3j + k + m(9i - 2j + 5k) are:$
(a) <2, -3, 1>
(b) <9, 2, 5>
(c) <-2, 3, -1>
(d) <9, -2, 5>
›Reveal solutionSolution

In the vector form $\vec r = \vec a + m\vec b$ , the coefficients of the parameter $m$ (i.e. $\vec b$ ) directly give the direction ratios of the line.

The line is $\vec r = 2\hat i - 3\hat j + \hat k + m(9\hat i - 2\hat j + 5\hat k)$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. Direction ratios of straight line r⃗ = 4î - 7ĵ + 5k̂ + s(9î - 2ĵ + 5k̂) are ______.$
›Reveal solutionSolution

In the vector form $\vec r = \vec a + s\,\vec b$ , the coefficients of the parameter give the line's direction ratios.

For $\vec r = 4\hat i - 7\hat j + 5\hat k + s(9\hat i - 2\hat j + 5\hat k)$ , the fixed point is $(4,-7,5)$ and the direction vector is …

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CBSE 2024Set ANNUAL1 mark
$(2, 3, -4)$
›Reveal solutionSolution

Direction ratios of a line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$ .

Let the points be $P(2,3,-4)$ and $Q(1,3,-2)$ .

Direction ratios of line PQ: …

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CBSE 2022Set ANNUAL1 markMCQ
$3x + 4y + 5z - 6 = 0$
(a) $3, 4, 5$
(b) $-3, 4, 5$
(c) $3, -4, 5$
(d) $2, 3, -4$
›Reveal solutionSolution

For $ax+by+cz+d=0$ , the normal has direction ratios $(a,b,c)=(3,4,5)$ .

The normal to a plane $ax+by+cz+d=0$ has direction ratios equal to the coefficients $(a,b,c)$ …

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CBSE 2019Set HE1 mark
$Q. Fill in the blank: Direction ratio of two parallel lines will be ______.$
›Reveal solutionSolution

Parallel lines have the same direction, so their direction ratios are proportional.

Direction ratios $(a,b,c)$ describe the direction a line points in. Two lines are parallel precisely when they point in the same (or exactly opposite) direction, which means one line's direction ratios are a scalar multiple of the oth …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Direction ratios of the line joining points A(2, 3, -4) and B(1, -2, 3) is:$
(a)
(i) -1, 5, 7
(b)
(ii) -1, -5, 7
(c)
(iii) 1, -5, -7
(d)
(iv) -1, -5, 7
›Reveal solutionSolution

Direction ratios $=(-1,-5,7)$ — option (iv).

Concept. For a line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ , the direction ratios are $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ .

Steps. With $A(2,3,-4)$ and $B(1,-2,3)$ : …

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Q.The vector having initial and terminal points as $(2, 5, 0)$ and $(-3, 7, 4)$ , respectively is
(A) $-\hat{i}+12\hat{j}+4\hat{k}$
(B) $5\hat{i}+2\hat{j}-4\hat{k}$
(C) $-5\hat{i}+2\hat{j}+4\hat{k}$
(D) $\hat{i}+\hat{j}+\hat{k}$

Concept understanding — Direction Vectors

Direction Vectors

The Idea

Vector Equation of a Line

Direction Ratios and Direction Cosines

Why It Matters …

Why Direction Vectors Work This Way

Step-by-Step

Method: Vector Joining Two Points (Terminal minus Initial)

Steps

Common Mistakes

Why Direction Vectors Work This Way

Step-by-Step

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