Sample 2026-27 Class 12 Maths Q14 Solution: Angle of i + j + √2 k with the positive z-axis

The question

Angle made by the vector $\hat i + \hat j + \sqrt{2}\,\hat k$ with positive direction of z-axis is

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{4}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{2}$

Exam answer

\begin{aligned} |\vec r| &= \sqrt{1^2 + 1^2 + (\sqrt{2})^2} \\ &= \sqrt{1 + 1 + 2} = 2 \end{aligned}

Why?

Call the vector $\vec r$ ⁠. Its magnitude is the square root of the sum of the squares of its components $1, 1, \sqrt{2}$ ⁠, and $(\sqrt{2})^2 = 2$ ⁠.

So the direction cosines are $\frac{1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}$ ⁠, that is $\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}}$ ⁠.

Why?

The direction cosines are the components divided by the magnitude:

l = \frac{1}{2},\ m = \frac{1}{2},\ n = \frac{\sqrt{2}}{2}

Also $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ ⁠. Check: $l^2 + m^2 + n^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1$ ⁠.

Write the 2 below as $\sqrt{2} \times \sqrt{2}$ and cancel one $\sqrt{2}$ ⁠:

\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{1}{\sqrt{2}}

The check, one square at a time:

\begin{aligned} \left(\tfrac{1}{2}\right)^2 &= \tfrac{1}{4},\quad \left(\tfrac{1}{\sqrt{2}}\right)^2 = \tfrac{1}{2} \\[4pt] \tfrac{1}{4} + \tfrac{1}{4} &= \tfrac{2}{4} = \tfrac{1}{2} \\ \tfrac{1}{2} + \tfrac{1}{2} &= 1 \end{aligned}

If $\gamma$ is the angle with the positive $z$ -axis, then $\cos\gamma = n$ ⁠:

\cos\gamma = \frac{1}{\sqrt{2}} \Rightarrow \gamma = \frac{\pi}{4}

Why?

The angle with an axis lies in $[0, \pi]$ ⁠, and in that interval only $\frac{\pi}{4}$ has cosine $\frac{1}{\sqrt{2}}$ ⁠.

\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}

Watch out: $\frac{\pi}{3}$ in (C) is the angle with the $x$ -axis (or the $y$ -axis), since $\cos\frac{\pi}{3} = \frac{1}{2}$ ⁠.

Answer: (B) $\dfrac{\pi}{4}$

Where this comes from

Chapter 10, Vector Algebra: Direction cosines of a vector.

Question type: The vector from A to B, and its direction cosines

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: The vector $\hat i + \hat j + \sqrt2\,\hat k$ is given directly, so skip the subtraction. Its length is 2, so the $\hat k$ direction cosine is $\frac{1}{\sqrt2}$ and $\gamma = \frac{\pi}{4}$ ⁠.

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Q14: Angle of i^+j^+2 k^\hat i + \hat j + \sqrt2\,\hat ki^+j^​+2​k^ with the positive z-axis

Q14: Angle of $\hat i + \hat j + \sqrt2\,\hat k$ with the positive z-axis