Let a = i + j + k , b =2 i +4 j -5 k and c =x i +2 j +3 k , x ∈ ℝ . If d is the unit vector in the direction of b + c such that a · d =1 , then ( a × b ) · c is equal to

Correct answer: A. 11 aligned & d =λ( b + c ) & a · d =λ( b · a + c · a ) & 1=λ(1+x+5) aligned 1=λ(x+6) ...(1) aligned & | d |=1 & |λ( b + c )|=1 & |λ((x+2) i +6 j -2 k )|=1 & λ^2((x+2)^2+6^2+2^2)=1 & x^2+4 x+4+36+4=(x+6)^2 & x^2+4 x+44=x^2+12 x+36 & 8 x=8, x=1 & | array ccc 1 & 1 & 1 2 & 4 & -5 x & 2 & 3 array |=( a × b ) · c & | array ccc 0 & 0 & 1 -2 & 9 & -4 x-2 & -1 & 3 array |=2-9(x-2) & =20-9 x & at x=1 & 20-9=11 aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Vector Algebra

Let a = i + j + k , b =2 i +4 j -5 k and c =x i +2 j +3 k , x ∈ ℝ ....

Let

\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}

and

\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}

. If

\vec{d}

is the unit vector in the direction of

\vec{b}+\vec{c}

such that

\vec{a} \cdot \vec{d}=1

, then

(\vec{a} \times \vec{b}) \cdot \vec{c}

is equal to

Options

Correct answer

A. 11

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \\ & 1=\lambda(1+\mathrm{x}+5)\end{aligned}

1=\lambda(x+6)

...(1)

\begin{aligned} & |\overrightarrow{\mathrm{d}}|=1 \\ & |\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1 \\ & |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 \\ & \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2\right)=1 \\ & \mathrm{x}^2+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^2 \\ & \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 \\ & 8 \mathrm{x}=8, \mathrm{x}=1 \\ & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & -5 \\ \mathrm{x} & 2 & 3 \end{array}\right|=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}} \\ & \left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 & 9 & -4 \\ \mathrm{x}-2 & -1 & 3 \end{array}\right|=2-9(\mathrm{x}-2) \\ & =20-9 \mathrm{x} \\ & \text{ at } \mathrm{x}=1 \\ & 20-9=11 \end{aligned}

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