Let three vectors a =α i +4 j +2 k , b =5 i +3 j +4 k , c =x i +y j +z k form a triangle such that c = a - b and the area of the triangle is 5 6 . If α is a positive real number, then | c |^2 is equal to:

Correct answer: B. 14 aligned & c = a - b & (x, y, z)=(α-5,1,-2) & x=α-5, y=1, z=-2.....(1) aligned Area of Δ=5 6 (given) aligned & (1)/(2)| a × c |=5 6 & \| array ccc i & j & k α & 4 & 2 x & 1 & -2 array \|=10 6 aligned aligned & |-10 i - j (-2 α-2 x)+ k (α-4 x)|=10 6 & (2 α+2 α-10)^2+(α-4 α+20)^2=500 & (4 α-10)^2+(20-3 α)^2=500 & 25 α^2-80 α-120 α=0 & α(25 α-200)=0 & α=8 (given α is +ve number) & x=α-5=3 & | c |^2=x^2+y^2+z^2 & =9+1+4 & =14 aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Vector Algebra

Let three vectors a =α i +4 j +2 k , b =5 i +3 j +4 k , c =x i +y j...

Let three vectors

\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}

form a triangle such that

\vec{c}=\vec{a}-\vec{b}

and the area of the triangle is

5 \sqrt{6}

. If

\alpha

is a positive real number, then

|\vec{c}|^2

is equal to:

Options

Correct answer

B. 14

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \\ & \Rightarrow(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\alpha-5,1,-2) \\ & \Rightarrow \mathrm{x}=\alpha-5, \mathrm{y}=1, \mathrm{z}=-2.....(1)\end{aligned}

Area of

\Delta=5 \sqrt{6}

(given)

\begin{aligned} & \frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=5 \sqrt{6} \\ & \left\|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \alpha & 4 & 2 \\ \mathrm{x} & 1 & -2\end{array}\right\|=10 \sqrt{6}\end{aligned}

\begin{aligned} & \Rightarrow|-10 \hat{\mathrm{i}}-\hat{\mathrm{j}}(-2 \alpha-2 \mathrm{x})+\hat{\mathrm{k}}(\alpha-4 \mathrm{x})|=10 \sqrt{6} \\ & \Rightarrow(2 \alpha+2 \alpha-10)^2+(\alpha-4 \alpha+20)^2=500 \\ & \Rightarrow(4 \alpha-10)^2+(20-3 \alpha)^2=500 \\ & \Rightarrow 25 \alpha^2-80 \alpha-120 \alpha=0 \\ & \Rightarrow \alpha(25 \alpha-200)=0 \\ & \Rightarrow \alpha=8 \text{ (given } \alpha \text{ is +ve number) } \\ & \Rightarrow \mathrm{x}=\alpha-5=3 \\ & |\overrightarrow{\mathrm{c}}|^2=\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 \\ & =9+1+4 \\ & =14\end{aligned}

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