GATE 2020 IN Q.1: solution and answer

GATE 2020 · INQ.1 · 1 mark · MCQ

The unit vectors along the mutually perpendicular $x$ , $y$ and $z$ axes are $\overset{}{^}$ , $\overset{}{^}$ and $\hat{k}$ respectively. Consider the plane $z = 0$ and two vectors $a$ and $b$ on that plane such that $a \neq = α b$ for any scalar $α$ . A vector perpendicular to both $a$ and $b$ is ____.

(A)
$\hat{k}$
(B)
$\overset{}{^} - \overset{}{^}$
(C)
$- \overset{}{^}$
(D)
$\overset{}{^}$

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Answer and solution

ANSWER

A vector perpendicular to both $a$ and $b$ is $\hat{k}$ , option (A).

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

Any vector in the plane $z = 0$ has zero $z$ -component, so write $a = a_{1} \overset{}{^} + a_{2} \overset{}{^}$ and $b = b_{1} \overset{}{^} + b_{2} \overset{}{^}$ .
The condition $a \neq = α b$ for every scalar $α$ means the two vectors are not parallel. So $a_{1} b_{2} - a_{2} b_{1} \neq = 0$ , and $a$ and $b$ together span the whole plane $z = 0$ .
The cross product $a \times b$ is perpendicular to both vectors: $a \times b = \overset{}{^} a_{1} b_{1} \overset{}{^} a_{2} b_{2} \hat{k} 00 = (a_{1} b_{2} - a_{2} b_{1}) \hat{k}$ Because $a_{1} b_{2} - a_{2} b_{1} \neq = 0$ , this is a nonzero multiple of $\hat{k}$ .
Direct check of option (A): $\hat{k} \cdot a = 0$ and $\hat{k} \cdot b = 0$ , since neither vector has a $z$ -component. So $\hat{k}$ , the normal to the plane $z = 0$ , is perpendicular to both.
Options (B), (C) and (D) ( $\overset{}{^} - \overset{}{^}$ , $- \overset{}{^}$ and $\overset{}{^}$ ) are nonzero vectors lying in the plane $z = 0$ . Suppose an in-plane vector $v$ were perpendicular to both $a$ and $b$ . Since $a$ and $b$ span the plane, $v$ would be perpendicular to every vector in the plane, including itself, so $v \cdot v = 0$ and $v = 0$ . So none of (B), (C), (D) can be perpendicular to both. For example, $\overset{}{^} \cdot a = a_{1}$ and $\overset{}{^} \cdot b = b_{1}$ , and these cannot both be zero, because then $a_{1} b_{2} - a_{2} b_{1} = 0$ .
So the answer is $\hat{k}$ , option (A).

KEY IDEA

Two non-parallel vectors in the plane $z = 0$ span that plane. The only direction perpendicular to both is the plane's normal, which is $\hat{k}$ . The cross product shows this: $a \times b = (a_{1} b_{2} - a_{2} b_{1}) \hat{k}$ .

WATCH OUT FOR

Picking an in-plane option such as $\overset{}{^}$ or $- \overset{}{^}$ because it is perpendicular to one particular vector you have in mind. The question asks for a vector perpendicular to BOTH of two non-parallel vectors in the plane, and no nonzero vector in the plane can be. Another slip is to overlook the condition $a \neq = α b$ . It is what makes $a \times b$ nonzero, so the perpendicular direction is exactly $\hat{k}$ .

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GATE 2020 IN Q.1: solution and answer

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