JEE Advanced 2023 Paper 2, Mathematics Section 1 Q4: Applications of Vectors

JEE Advanced2023Paper 2Mathematics Section 1

Let the position vectors of the points P, Q, R and S be $a = \hat{i} + 2 \hat{j} - 5 \hat{k}$ , $b = 3 \hat{i} + 6 \hat{j} + 3 \hat{k}$ , $c = \frac{17}{5} \hat{i} + \frac{16}{5} \hat{j} + 7 \hat{k}$ and $d = 2 \hat{i} + \hat{j} + \hat{k}$ , respectively. Then which of the following statements is true?

A
The points P, Q, R and S are NOT coplanar
B
$\frac{b + 2 d}{3}$ is the position vector of a point which divides PR internally in the ratio $5 : 4$
C
$\frac{b + 2 d}{3}$ is the position vector of a point which divides PR externally in the ratio $5 : 4$
D
The square of the magnitude of the vector $b \times d$ is $95$

Solution

Compute $\frac{b + 2 d}{3} = \frac{( 3 i ^ + 6 j ^ + 3 k ^ ) + ( 4 i ^ + 2 j ^ + 2 k ^ )}{3} = \frac{7 i ^ + 8 j ^ + 5 k ^}{3} .$

Now check the point $\frac{5 c + 4 a}{9}$ . We have $5 c = 17 \hat{i} + 16 \hat{j} + 35 \hat{k}$ and $4 a = 4 \hat{i} + 8 \hat{j} - 20 \hat{k}$ , so $\frac{5 c + 4 a}{9} = \frac{21 i ^ + 24 j ^ + 15 k ^}{9} = \frac{7 i ^ + 8 j ^ + 5 k ^}{3} .$

Hence $\frac{b + 2 d}{3} = \frac{5 c + 4 a}{9}$ , which is precisely the section-formula point dividing PR (from P with position $a$ to R with position $c$ ) internally in the ratio $5 : 4$ . Option (B) is correct.

For option (D): $∣ b \times d ∣^{2} = ∣ b ∣^{2} ∣ d ∣^{2} - (b \cdot d)^{2} = 54 \cdot 6 - 1 5^{2} = 324 - 225 = 99 \neq = 95.$

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JEE Advanced 2023 Paper 2, Mathematics Section 1 Q4: Applications of Vectors

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