Let a,b and c be distinct positive numbers. If the vectors ai^+aj^+ck^,i^+k^ and ci^+cj^+bk^ are co-planar, then c is equal to:

Correct answer: (D) √ab. Given vectors ai^+aj^+ck^,i^+k^ and ci^+cj^+bk^ are co-planar, then a a c 1 0 1 c c b =0 ⇒a0-c-ab-c+cc-0=0 ⇒c^2=ab ⇒c=√ab.

Q39 / 90 · JEE Main 2021 (25 Jul Shift 2) · Mathematics

Let $a, b$ and $c$ be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are co-planar, then $c$ is equal to:

Given vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are co-planar, then

$|\begin{array}{l} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{array}| = 0$

$\Rightarrow a (0 - c) - a (b - c) + c (c - 0) = 0$

$\Rightarrow c^{2} = a b$

$\Rightarrow c = \sqrt{a b} .$

Hence, the correct answer is option: 'D'.