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Vector Algebra question

2024 · 6 Apr · Shift 1 · Q86
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Vector Algebra question

2024 · 6 Apr · Shift 1 · Q86

JEE MainPhysicsVector AlgebraNumerical+4 / −1
For three vectors A⃗=(−xi^−6j^−2k^),B⃗=(−i^+4j^+3k^)\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k}), \vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})A=(−xi^−6j^​−2k^),B=(−i^+4j^​+3k^) and C⃗=(−8i^−j^+3k^)\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k})C=(−8i^−j^​+3k^), if A⃗⋅(B⃗×C⃗)=0\vec{A} \cdot(\vec{B} \times \vec{C})=0A⋅(B×C)=0, then value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

  1. We are given:
A⃗=(−xi^−6j^−2k^),B⃗=(−i^+4j^+3k^),C⃗=(−8i^−j^+3k^)\vec A = (-x\hat i-6\hat j-2\hat k),\qquad \vec B = (-\hat i+4\hat j+3\hat k),\qquad \vec C = (-8\hat i-\hat j+3\hat k)A=(−xi^−6j^​−2k^),B=(−i^+4j^​+3k^),C=(−8i^−j^​+3k^)

So in component form:

A⃗=(−x,−6,−2),B⃗=(−1,4,3),C⃗=(−8,−1,3)\vec A = (-x,-6,-2),\quad \vec B = (-1,4,3),\quad \vec C = (-8,-1,3)A=(−x,−6,−2),B=(−1,4,3),C=(−8,−1,3)
  1. Since
A⃗⋅(B⃗×C⃗)=0\vec A\cdot(\vec B\times \vec C)=0A⋅(B×C)=0

this is the scalar triple product. We compute it using the determinant:

A⃗⋅(B⃗×C⃗)=∣−x−6−2−143−8−13∣=0\vec A\cdot(\vec B\times \vec C)= \begin{vmatrix} -x & -6 & -2\\ -1 & 4 & 3\\ -8 & -1 & 3 \end{vmatrix}=0A⋅(B×C)=​−x−1−8​−64−1​−233​​=0
  1. Expand along the first row:
∣−x−6−2−143−8−13∣=(−x)∣43−13∣−(−6)∣−13−83∣+(−2)∣−14−8−1∣\begin{vmatrix} -x & -6 & -2\\ -1 & 4 & 3\\ -8 & -1 & 3 \end{vmatrix} = (-x) \begin{vmatrix} 4 & 3\\ -1 & 3 \end{vmatrix} -(-6) \begin{vmatrix} -1 & 3\\ -8 & 3 \end{vmatrix} +(-2) \begin{vmatrix} -1 & 4\\ -8 & -1 \end{vmatrix}​−x−1−8​−64−1​−233​​=(−x)​4−1​33​​−(−6)​−1−8​33​​+(−2)​−1−8​4−1​​
  1. Now compute each minor:
∣43−13∣=4⋅3−3(−1)=12+3=15\begin{vmatrix} 4 & 3\\ -1 & 3 \end{vmatrix} = 4\cdot 3 - 3(-1)=12+3=15​4−1​33​​=4⋅3−3(−1)=12+3=15 ∣−13−83∣=(−1)(3)−3(−8)=−3+24=21\begin{vmatrix} -1 & 3\\ -8 & 3 \end{vmatrix} = (-1)(3)-3(-8)=-3+24=21​−1−8​33​​=(−1)(3)−3(−8)=−3+24=21 ∣−14−8−1∣=(−1)(−1)−4(−8)=1+32=33\begin{vmatrix} -1 & 4\\ -8 & -1 \end{vmatrix} = (-1)(-1)-4(-8)=1+32=33​−1−8​4−1​​=(−1)(−1)−4(−8)=1+32=33
  1. Substitute back:
(−x)(15)−(−6)(21)+(−2)(33)=0(-x)(15)-(-6)(21)+(-2)(33)=0(−x)(15)−(−6)(21)+(−2)(33)=0 −15x+126−66=0-15x+126-66=0−15x+126−66=0 −15x+60=0-15x+60=0−15x+60=0 15x=6015x=6015x=60 x=4x=4x=4
  1. Therefore, the required integer value is:
4\boxed{4}4​
  1. Comparison with stored answer:

Stored correct answer = 444, which matches our result.

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