CBSE Board Examination 2025 · Section A · 1 mark

Q2: Which is true for $\vec a = 3\hat i + 2\hat j - \hat k$ and $\vec b = \hat i - \hat j + \hat k$ ⁠?

The question

If vector $\vec a = 3\hat i + 2\hat j - \hat k$ and vector $\vec b = \hat i - \hat j + \hat k$ ⁠, then which of the following is correct ?

(A) $\vec a \parallel \vec b$

(B) $\vec a \perp \vec b$

(D) $|\vec a| = |\vec b|$

Exam answer

\begin{aligned} \vec a \cdot \vec b &= (3)(1) + (2)(-1) \\ &\quad + (-1)(1) \\ &= 3 - 2 - 1 = 0 \end{aligned}

Why?

In components, the scalar (dot) product multiplies matching components and adds:

\begin{aligned} \vec a \cdot \vec b &= a_1b_1 + a_2b_2 \\ &\quad + a_3b_3 \end{aligned}

Here the components of $\vec a$ are $3, 2, -1$ and those of $\vec b$ are $1, -1, 1$ ⁠.

$\vec a \cdot \vec b = 0$ ⁠, so $\vec a \perp \vec b$ ⁠.

Why?

The dot product is also

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

Neither vector is zero, so $\vec a \cdot \vec b = 0$ forces $\cos\theta = 0$ ⁠, that is, $\theta = \frac{\pi}{2}$ ⁠.

\begin{aligned} |\vec a| &= \sqrt{9 + 4 + 1} = \sqrt{14} \\ |\vec b| &= \sqrt{1 + 1 + 1} = \sqrt{3} \end{aligned}

$|\vec b| < |\vec a|$ ⁠, so (C) and (D) are false.

$\frac{3}{1} \ne \frac{2}{-1}$ ⁠, so $\vec a$ is not parallel to $\vec b$ ⁠: (A) is false.

Why?

The magnitude is $\sqrt{a_1^2 + a_2^2 + a_3^2}$ ⁠. Square each component, with brackets round the negatives:

\begin{aligned} |\vec a| &= \sqrt{(3)^2 + (2)^2 + (-1)^2} \\ &= \sqrt{9 + 4 + 1} = \sqrt{14} \\[4pt] |\vec b| &= \sqrt{(1)^2 + (-1)^2 + (1)^2} \\ &= \sqrt{1 + 1 + 1} = \sqrt{3} \end{aligned}

Since $14 > 3$ ⁠, $\sqrt{14} > \sqrt{3}$ ⁠.

Parallel vectors are multiples of each other, so their components are in one fixed ratio. Here $\frac{3}{1} = 3$ but $\frac{2}{-1} = -2$ ⁠.

Watch out: a zero dot product means perpendicular, not parallel.

Answer: (B) $\vec a \perp \vec b$

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: Option (B) says $\vec a \perp \vec b$ ⁠, so test it with the dot product: $3 - 2 - 1 = 0$ ⁠, and the vectors are perpendicular.

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Vector Algebra in past papers: question 6 of 48 · see them all

Q2: Which is true for a⃗=3i^+2j^−k^\vec a = 3\hat i + 2\hat j - \hat ka=3i^+2j^​−k^ and b⃗=i^−j^+k^\vec b = \hat i - \hat j + \hat kb=i^−j^​+k^⁠?