Board 2026 Class 12 Maths Q23 Solution: Vector of magnitude 14 in the direction of QP

The question

Find the vector of magnitude 14 in the direction of $\overrightarrow{QP}$ ⁠, where P and Q are the points $(1, 3, 2)$ and $(-1, 0, 8)$ respectively.

Exam answer

\begin{aligned} \overrightarrow{QP} &= (1 + 1)\hat i + (3 - 0)\hat j \\ &\quad + (2 - 8)\hat k \\ &= 2\hat i + 3\hat j - 6\hat k \end{aligned}

Why?

A vector from one point to another is "head minus tail":

\overrightarrow{QP} = \vec p - \vec q

Here the head is $P(1, 3, 2)$ and the tail is $Q(-1, 0, 8)$ ⁠, so subtract the coordinates of Q from those of P.

Watch out: $1 - (-1) = 1 + 1 = 2$ ⁠. Taking P from Q instead would give the opposite direction.

\begin{aligned} |\overrightarrow{QP}| &= \sqrt{2^2 + 3^2 + (-6)^2} \\ &= \sqrt{4 + 9 + 36} = 7 \end{aligned}

Why?

The magnitude of $a\hat i + b\hat j + c\hat k$ is

\sqrt{a^2 + b^2 + c^2}

Here $4 + 9 + 36 = 49$ and $\sqrt{49} = 7$ ⁠.

Unit vector in the direction of $\overrightarrow{QP}$ ⁠:

\frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} = \frac27\hat i + \frac37\hat j - \frac67\hat k

Why?

Dividing a vector by its own magnitude keeps its direction and makes its length 1. That is the unit vector.

Required vector $= 14\,\frac{\overrightarrow{QP}}{|\overrightarrow{QP}|}$

= 14\left(\frac27\hat i + \frac37\hat j - \frac67\hat k\right)

Why?

A unit vector has length 1, so 14 times it has length 14 and the same direction.

Multiply each component by 14: $14\cdot\frac27 = 4$ ⁠, $14\cdot\frac37 = 6$ and $14\cdot\frac67 = 12$ ⁠.

\begin{aligned} 14\cdot\frac27 &= \frac{28}{7} = 4 \\ 14\cdot\frac37 &= \frac{42}{7} = 6 \\ 14\cdot\left(-\frac67\right) &= -\frac{84}{7} = -12 \end{aligned}

$\therefore$ Required vector $= 4\hat i + 6\hat j - 12\hat k$ ⁠.

Where this comes from

Chapter 10, Vector Algebra: Vector of given magnitude in a given direction.

Question type: The vector from A to B, and its direction cosines + A unit vector, or a vector of a given length (combines 2 types)

How to spot it: click to see

Type 2. The vector from A to B, and its direction cosines

Spot it: two points are given, like $A(1, 2, -3)$ and $B(-1, -2, 1)$ ⁠.

Here: First, the points $P(1, 3, 2)$ and $Q(-1, 0, 8)$ ⁠: $\overrightarrow{QP}$ is P minus Q, $2\hat i + 3\hat j - 6\hat k$ ⁠, with length 7.

Type 1. A unit vector, or a vector of a given length

Spot it: "unit vector in the direction of", or "a vector of magnitude 8 in the direction of".

Here: Then the magnitude 14: divide $\overrightarrow{QP}$ by 7 and multiply by 14, giving $4\hat i + 6\hat j - 12\hat k$ ⁠.

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q23: Vector of magnitude 14 in the direction of QP→\overrightarrow{QP}QP​