CBSE Board Examination 2026 · Section A · 1 mark

Q13: Value of $p$ for two perpendicular vectors

The question

The value of p for which vectors $\hat i + 2\hat j + 3\hat k$ and $2\hat i - p\hat j + \hat k$ are perpendicular to each other is

(A) $0$

(B) $1$

(D) $-\frac{5}{2}$

Exam answer

Perpendicular vectors have scalar product 0.

Why?

The scalar product uses the angle $\theta$ between the vectors:

\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta

For perpendicular vectors $\theta = 90^\circ$ and $\cos 90^\circ = 0$ ⁠, so the product is 0.

\begin{aligned} &(1)(2) + (2)(-p) + (3)(1) = 0 \\ \Rightarrow\ &2 - 2p + 3 = 0 \\ \Rightarrow\ &5 - 2p = 0 \Rightarrow p = \frac{5}{2} \end{aligned}

Why?

In components, multiply matching components and add:

\begin{aligned} &(a_1\hat i + a_2\hat j + a_3\hat k) \\ &\cdot (b_1\hat i + b_2\hat j + b_3\hat k) \\ &= a_1b_1 + a_2b_2 + a_3b_3 \end{aligned}

Then solve, one move at a time:

\begin{aligned} 2 + 3 - 2p &= 0 \\ 5 - 2p &= 0 \\ 5 &= 2p \\ p &= \frac{5}{2} \end{aligned}

Watch out: the $\hat j$ component of the second vector is $-p$ ⁠, not $p$ ⁠. Using $+p$ gives $p = -\frac{5}{2}$ ⁠, option (D).

Answer: (C) $\dfrac{5}{2}$

Where this comes from

Chapter 10, Vector Algebra: Scalar product: perpendicular vectors.

Question type: Perpendicular vectors

How to spot it: click to see

Type 5. Perpendicular vectors: set the dot product to 0

Spot it: "is perpendicular to", with an unknown such as $\lambda$ to find.

Here: The unknown is the $-p$ in the second vector's $\hat j$ part. Set the dot product $2 - 2p + 3$ equal to 0, so $p = \frac52$ ⁠.

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 1 of 48 · see them all

Q13: Value of ppp for two perpendicular vectors