Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be three points whose position vectors respectively are: $\vec{a}=\hat{i}+4 \hat{j}+3 \hat{k}$ $\vec{b}=2 \hat{i}+\alpha \hat{j}+4 \hat{k}, \alpha \in \mathbb{R}$ $\vec{c}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, in $\triangle \mathrm{ABC}$, through A is:

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Vector Algebra

Let

A, B, C

be three points whose position vectors respectively are:

a = \hat{i} + 4 \hat{j} + 3 \hat{k}

b = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R

c = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}

α

is the smallest positive integer for which

a, b, c

are non-collinear, then the length of the median, in

△ ABC

, through A is:

Answer: (A)

For collinearity,

A B ∥ A C

. We have

A B = \hat{i} + (α - 4) \hat{j} + \hat{k}

and

A C = 2 \hat{i} - 6 \hat{j} + 2 \hat{k}

.

For parallel vectors:

\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2}

, giving

α = 1

.

The smallest positive integer for which the points are non-collinear is

α = 2

.

With

α = 2

A (1, 4, 3)

B (2, 2, 4)

C (3, - 2, 5)

.

The midpoint of

B C

M (\frac{5}{2}, 0, \frac{9}{2})

.

The median length is:

A M = (\frac{5}{2} - 1)^{2} + (0 - 4)^{2} + (\frac{9}{2} - 3)^{2} = \frac{9}{4} + 16 + \frac{9}{4} = \frac{82}{2}

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option