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JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Vector Algebra

Let

a = α \hat{i} + \hat{j} - \hat{k}

and

b = 2 \hat{i} + \hat{j} - α \hat{k}, α > 0

. If the projection of

a \times b

on the vector

- \hat{i} + 2 \hat{j} - 2 \hat{k}

is 30, then

α

is equal to

(A) $\frac{15}{2}$
(B) $8$
(C) $\frac{13}{2}$
(D) $7$ ✓ Correct

Answer: (D)

Solution

First compute

a \times b = \hat{i} α 2 \hat{j} 11 \hat{k} - 1 - α = \hat{i} (- α + 1) - \hat{j} (- α^{2} + 2) + \hat{k} (α - 2) = (1 - α) \hat{i} + (α^{2} - 2) \hat{j} + (α - 2) \hat{k}

. The projection on

c = - \hat{i} + 2 \hat{j} - 2 \hat{k}

is

\frac{( a \times b ) \cdot c}{∣ c ∣} = \frac{- ( 1 - α ) + 2 ( α ^{2} - 2 ) - 2 ( α - 2 )}{3} = \frac{- 1 + α + 2 α ^{2} - 4 - 2 α + 4}{3} = \frac{2 α ^{2} - α - 1}{3} = 30

. Thus

2 α^{2} - α - 1 = 90 \Rightarrow 2 α^{2} - α - 91 = 0

. Using the quadratic formula:

α = \frac{1 \pm 1 + 728}{4} = \frac{1 \pm 27}{4}

. Since

α > 0

, we have

α = 7

.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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