JEE Advanced 2022 Paper 2, Mathematics Section 2 Q5: Scalar and Vector Triple Product

JEE Advanced2022Paper 2Mathematics Section 2

Let $\hat{i}, \hat{j}$ , and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let

$a = 3 \hat{i} + \hat{j} - \hat{k}$ ,

$b = \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ , $b_{2}, b_{3} \in R$ ,

$c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ , $c_{1}, c_{2}, c_{3} \in R$

be three vectors such that $b_{2} b_{3} > 0$ , $a \cdot b = 0$ and

$0 - c_{3} c_{2} c_{3} 0 - c_{1} - c_{2} c_{1} 0 1 b_{2} b_{3} = 3 - c_{1} 1 - c_{2} - 1 - c_{3}$ .

Then, which of the following is/are TRUE?

A
$a \cdot c = 0$
B
$b \cdot c = 0$
C
$∣ b ∣ > 10$
D
$∣ c ∣ \leq 11$

Solution

The matrix equation is $c \times b = a - c$ (the cross-product representation).

Dot with $b$ : $(c \times b) \cdot b = 0 = a \cdot b - c \cdot b$ . Given $a \cdot b = 0$ , this yields $b \cdot c = 0$ . Option (B) holds.

Dot with $c$ : $(c \times b) \cdot c = 0 = a \cdot c - ∣ c ∣^{2}$ , so $∣ c ∣^{2} = a \cdot c = ∣ a ∣∣ c ∣ cos ϕ$ for some angle $ϕ$ . Hence $∣ c ∣ \leq ∣ a ∣ = 9 + 1 + 1 = 11$ . Option (D) holds; option (A) fails in general (it would require $∣ c ∣ = 0$ ).

From $a \cdot b = 0$ : $3 + b_{2} - b_{3} = 0$ , so $b_{3} - b_{2} = 3$ . Then $∣ b ∣^{2} = 1 + b_{2}^{2} + b_{3}^{2} = 1 + (b_{3} - b_{2})^{2} + 2 b_{2} b_{3} = 10 + 2 b_{2} b_{3}$ . Since $b_{2} b_{3} > 0$ , $∣ b ∣^{2} > 10$ , i.e. $∣ b ∣ > 10$ . Option (C) holds.

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JEE Advanced 2022 Paper 2, Mathematics Section 2 Q5: Scalar and Vector Triple Product

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