Let A, B, C be three points whose position vectors respectively are: → a = ^ i + 4 ^ j + 3 ^ k → b = 2 ^ i + α ^ j + 4 ^ k , α ∈ R → c = 3 ^ i - 2 ^ j + 5 ^ k If α is the smallest positive integer for which → a , → b

Correct answer: A. 82 2 Let us assume that vectors are collinear so, → AB ∥ → AC if (1)/(2) = (α - 4)/(- 6) = (1)/(2) ⇒ α = 1 Now, for → a , → b , → c to be non-collinear smallest positive integer will be α = 2 So, mid-point of BC = M((5)/(2), 0, (9)/(2)) Now, AM = (9)/(4) + 16 + (9)/(4) = 82 2

JEE Main 202229 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2022 · 29 June · Evening Shift · Mathematics · Vector Algebra

Let A, B, C be three points whose position vectors respectively...

Let $A, B, C$ be three points whose position vectors respectively are:

$\overset{\to}{a} = \overset{^}{i} + 4\overset{^}{j} + 3\overset{^}{k}$

$\overset{\to}{b} = 2\overset{^}{i} + \alpha\overset{^}{j} + 4\overset{^}{k}, \alpha \in R$

$\overset{\to}{c} = 3\overset{^}{i} - 2\overset{^}{j} + 5\overset{^}{k}$

If $\alpha$ is the smallest positive integer for which $\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c}$ are non-collinear, then the length of the median, $△ ABC$ , through $A$ is:

Options

A $\dfrac{\sqrt{82}}{2}$
B $\dfrac{\sqrt{62}}{2}$
C $\dfrac{\sqrt{69}}{2}$
D $\dfrac{\sqrt{66}}{2}$

Correct answer

A. $\dfrac{\sqrt{82}}{2}$

Step-by-step solution

Let us assume that vectors are collinear so, $\overset{\to}{AB} ∥ \overset{\to}{AC}$ if $\dfrac{1}{2} = \dfrac{\alpha - 4}{- 6} = \dfrac{1}{2} ⇒ \alpha = 1$

Now, for $\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c}$ to be non-collinear smallest positive integer will be $\alpha = 2$

So, mid-point of $BC = M\left(\dfrac{5}{2}, 0, \dfrac{9}{2}\right)$

Now, $AM = \sqrt{\dfrac{9}{4} + 16 + \dfrac{9}{4}} = \dfrac{\sqrt{82}}{2}$

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