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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Vector Algebra

Let

a = α \hat{i} + 3 \hat{j} - \hat{k}, b = 3 \hat{i} - β \hat{j} + 4 \hat{k}

and

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

where

α, β \in R

, be three vectors. If the projection of

a

on

c

is

\frac{10}{3}

and

b \times c = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k}

, then the value of

α + β

is equal to :

(A)3✓ Correct
(B)4
(C)5
(D)6

Answer: (A)

Solution

The projection of

a

on

c

is:

\frac{a \cdot c}{∣ c ∣} = \frac{10}{3}

a \cdot c = α + 6 + 2 = α + 8

∣ c ∣ = 1 + 4 + 4 = 3

So:

\frac{α + 8}{3} = \frac{10}{3} \Rightarrow α = 2

For the cross product:

b \times c = \hat{i} 31 \hat{j} - β 2 \hat{k} 4 - 2 = (2 β - 8) \hat{i} + 10 \hat{j} + (6 + β) \hat{k}

Comparing with

- 6 \hat{i} + 10 \hat{j} + 7 \hat{k}

:

2 β - 8 = - 6 \Rightarrow β = 1

Therefore:

α + β = 3

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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