CBSE Sample Paper 2026-27 · Section A · 1 mark

Q13: Value of $u$ when $2\hat i - u\hat j + 3\hat k$ has magnitude 4

The question

If vector $2\hat i - u\hat j + 3\hat k$ has magnitude 4, then value of ‘ $u$ ’ is

(A) $1$

(B) $\sqrt{2}$

(D) $\pm 2$

Exam answer

The components are $2, -u, 3$ ⁠, and the magnitude is 4:

\begin{aligned} &\sqrt{2^2 + (-u)^2 + 3^2} = 4 \\ \Rightarrow\ &\sqrt{2^2 + u^2 + 3^2} = 4 \end{aligned}

Why?

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

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

Here $b = -u$ ⁠, and $(-u)^2 = u^2$ ⁠, so the sign of $u$ disappears.

\begin{aligned} \Rightarrow\ &4 + u^2 + 9 = 16 \\ \Rightarrow\ &u^2 = 16 - 13 \\ \Rightarrow\ &u^2 = 3 \Rightarrow u = \pm\sqrt{3} \end{aligned}

Why?

Square both sides to remove the root: $4^2 = 16$ ⁠. Then $4 + 9 = 13$ moves to the right.

\begin{aligned} 2^2 &= 4,\quad 3^2 = 9 \\[4pt] \left(\sqrt{4 + u^2 + 9}\right)^2 &= 4^2 \\ 4 + u^2 + 9 &= 16 \\ u^2 + 13 &= 16 \\ u^2 &= 16 - 13 = 3 \\ u &= \pm\sqrt{3} \end{aligned}

Both roots work, because only $u^2$ appears: $4 + 3 + 9 = 16$ and $\sqrt{16} = 4$ ⁠. Watch out: $\sqrt{3}$ alone misses $-\sqrt{3}$ ⁠.

Answer: (C) $\pm\sqrt{3}$

Where this comes from

Chapter 10, Vector Algebra: Magnitude of a vector.

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Vector Algebra in past papers: question 21 of 48 · see them all

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