CBSE Sample Paper 2026-27 · Section B · 2 marks

Q24: Find $|\vec b|$ from $|\vec a|$ ⁠, $\vec a \cdot \vec b$ and $\vec b \times \vec a$

Sample 2026-27Ch 10 · Vector Algebra

The question

For two non-zero vectors $\vec a$ and $\vec b$ ⁠, $|\vec a| = \sqrt6$ ⁠, $\vec a \cdot \vec b = 1$ and $\vec b \times \vec a = 3\hat i + 4\hat j - 2\hat k$ ⁠. Find $|\vec b|$ ⁠.

Exam answer

\begin{aligned} |\vec a \times \vec b| &= |\vec b \times \vec a| \\ &= \sqrt{9 + 16 + 4} = \sqrt{29} \end{aligned}

Why?

Swapping the order of a cross product only changes its sign, $\vec a \times \vec b = -(\vec b \times \vec a)$ ⁠, so the two have the same magnitude.

The magnitude of $3\hat i + 4\hat j - 2\hat k$ is

\begin{aligned} &\sqrt{3^2 + 4^2 + (-2)^2} \\ &= \sqrt{9 + 16 + 4} \\ &= \sqrt{25 + 4} = \sqrt{29} \end{aligned}

We know $|\vec a \times \vec b|^2 + |\vec a \cdot \vec b|^2$ $= |\vec a|^2|\vec b|^2$ ⁠.

Why?

This identity comes from the two products with the angle $\theta$ between the vectors:

\begin{aligned} |\vec a \times \vec b| &= |\vec a||\vec b|\sin\theta \\ \vec a \cdot \vec b &= |\vec a||\vec b|\cos\theta \end{aligned}

Square both and add. Since $\sin^2\theta + \cos^2\theta = 1$ ⁠, the sum is $|\vec a|^2|\vec b|^2$ ⁠.

\begin{aligned} \Rightarrow 29 + 1 &= 6|\vec b|^2 \\ \Rightarrow |\vec b|^2 &= \frac{30}{6} = 5 \end{aligned}

Why?

Substitute: $(\sqrt{29})^2 = 29$ ⁠, $1^2 = 1$ and $(\sqrt6)^2 = 6$ ⁠.

\begin{aligned} (\sqrt{29})^2 + (1)^2 &= (\sqrt6)^2|\vec b|^2 \\ 29 + 1 &= 6|\vec b|^2 \\ 30 &= 6|\vec b|^2 \\ |\vec b|^2 &= \frac{30}{6} = 5 \\ |\vec b| &= \sqrt5 \end{aligned}

A magnitude is never negative, so take the positive square root of 5.

$\therefore\ |\vec b| = \sqrt5$

Where this comes from

Chapter 10, Vector Algebra: Dot and cross product: finding a magnitude.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

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

Vector Algebra in past papers: question 24 of 48 · see them all

Q24: Find ∣b⃗∣|\vec b|∣b∣ from ∣a⃗∣|\vec a|∣a∣⁠, a⃗⋅b⃗\vec a \cdot \vec ba⋅b and b⃗×a⃗\vec b \times \vec ab×a

Q24: Find $|\vec b|$ from $|\vec a|$ ⁠, $\vec a \cdot \vec b$ and $\vec b \times \vec a$