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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Vector Algebra

Let

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

and

c = - α \hat{i} - 2 \hat{j} + \hat{k}

, where

α

and

β

are integers. If

a \cdot b = - 1

and

b \cdot c = 10

, then

(a \times b) \cdot c

is equal to ______

Answer: 9

Solution

a \cdot b = 3 - α β - α β = 3 - 2 α β = - 1 \Rightarrow α β = 2

.

b \cdot c = - 3 α - 2 β - α = - 4 α - 2 β = 10 \Rightarrow 2 α + β = - 5

.

From

α β = 2

and

2 α + β = - 5

:

β = - 5 - 2 α

.

Substituting:

α (- 5 - 2 α) = 2 \Rightarrow - 5 α - 2 α^{2} = 2 \Rightarrow 2 α^{2} + 5 α + 2 = 0

.

Solving:

(2 α + 1) (α + 2) = 0 \Rightarrow α = - 2

(integer),

β = - 1

.

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

,

b = 3 \hat{i} - \hat{j} + 2 \hat{k}

,

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

.

(a \times b) \cdot c

is the scalar triple product, which equals the determinant:

132 2 - 1 - 2 - 1 21 = 1 (- 1 + 4) - 2 (3 - 4) - 1 (- 6 + 2) = 3 + 2 + 4 = 9

.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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