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

Q15: A vector perpendicular to i^+j^\hat i + \hat j and i^−j^\hat i - \hat j

Sample 2026-27Ch 10 · Vector Algebra
The question

A vector perpendicular to the vectors i^+j^\hat i + \hat j and i^−j^\hat i - \hat j is

(A) −2j^-2\hat j

(B) 3k^3\hat k

(C) 2i^2\hat i

(D) 2i^+k^2\hat i + \hat k

Exam answer

Let a⃗=i^+j^\vec a = \hat i + \hat j and b⃗=i^−j^\vec b = \hat i - \hat j⁠. A vector perpendicular to both is along a⃗×b⃗\vec a \times \vec b⁠:

a⃗×b⃗=∣i^j^k^1101−10∣=i^(0−0)−j^(0−0)+k^(−1−1)\begin{aligned} \vec a \times \vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} \\ &= \hat i(0 - 0) - \hat j(0 - 0) \\ &\quad + \hat k(-1 - 1) \end{aligned} (i^+j^)×(i^−j^)=−2k^(\hat i + \hat j) \times (\hat i - \hat j) = -2\hat k
Why?

a⃗×b⃗\vec a \times \vec b is perpendicular to both a⃗\vec a and b⃗\vec b⁠. Expand the determinant along its first row. The k^\hat k part is

(1)(−1)−(1)(1)=−2(1)(-1) - (1)(1) = -2

and the i^\hat i⁠, j^\hat j parts are 0, because both vectors have k^\hat k component 0.

i^:(1)(0)−(0)(−1)=0−0=0j^:(1)(0)−(0)(1)=0−0=0k^:(1)(−1)−(1)(1)=−1−1=−2\begin{aligned} \hat i&: (1)(0) - (0)(-1) = 0 - 0 = 0 \\ \hat j&: (1)(0) - (0)(1) = 0 - 0 = 0 \\ \hat k&: (1)(-1) - (1)(1) = -1 - 1 = -2 \end{aligned}

So any vector along k^\hat k is perpendicular to both (both vectors lie in the XY-plane). Hence 3k^3\hat k⁠.

3k^⋅(i^+j^)=03k^⋅(i^−j^)=0−2j^⋅(i^+j^)=−2≠02i^⋅(i^+j^)=2≠0(2i^+k^)⋅(i^+j^)=2≠0\begin{aligned} 3\hat k \cdot (\hat i + \hat j) &= 0 \\ 3\hat k \cdot (\hat i - \hat j) &= 0 \\ -2\hat j \cdot (\hat i + \hat j) &= -2 \ne 0 \\ 2\hat i \cdot (\hat i + \hat j) &= 2 \ne 0 \\ (2\hat i + \hat k) \cdot (\hat i + \hat j) &= 2 \ne 0 \end{aligned}
Why?

Any multiple of −2k^-2\hat k⁠, such as 3k^3\hat k⁠, points along the zz-axis, and so is perpendicular to every vector in the XY-plane.

Two vectors are perpendicular exactly when their dot product is 0. Only 3k^3\hat k gives 0 with both, so the other options fail.

Multiply matching components (i^\hat i⁠, j^\hat j⁠, k^\hat k⁠) and add:

3k^⋅(i^+j^)=(0)(1)+(0)(1)+(3)(0)=03k^⋅(i^−j^)=(0)(1)+(0)(−1)+(3)(0)=0−2j^⋅(i^+j^)=(0)(1)+(−2)(1)+(0)(0)=−22i^⋅(i^+j^)=(2)(1)+(0)(1)+(0)(0)=2(2i^+k^)⋅(i^+j^)=(2)(1)+(0)(1)+(1)(0)=2\begin{aligned} &3\hat k \cdot (\hat i + \hat j) \\ &= (0)(1) + (0)(1) + (3)(0) = 0 \\[4pt] &3\hat k \cdot (\hat i - \hat j) \\ &= (0)(1) + (0)(-1) + (3)(0) = 0 \\[4pt] &-2\hat j \cdot (\hat i + \hat j) \\ &= (0)(1) + (-2)(1) + (0)(0) = -2 \\[4pt] &2\hat i \cdot (\hat i + \hat j) \\ &= (2)(1) + (0)(1) + (0)(0) = 2 \\[4pt] &(2\hat i + \hat k) \cdot (\hat i + \hat j) \\ &= (2)(1) + (0)(1) + (1)(0) = 2 \end{aligned}

Answer: (B) 3k^3\hat k

Where this comes from

Chapter 10, Vector Algebra: Vector perpendicular to two vectors.

Question type: A unit vector perpendicular to two vectors

How to spot it: click to see

Type 10. A unit vector perpendicular to two vectors

Spot it: "a unit vector perpendicular to each of the vectors".

Here: A vector perpendicular to i^+j^\hat i + \hat j and i^−j^\hat i - \hat j lies along their cross product, −2k^-2\hat k⁠. No unit vector is asked, so any multiple such as 3k^3\hat k works.

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