Mathematics in Physics

JEE Main level, numerical answer, medium

41 of 121

Question 41 of 121

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}) and C⃗=(−8i^−j^+3k^)\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k}), if A⃗⋅(B⃗×C⃗)=0\vec{A} \cdot(\vec{B} \times \vec{C})=0, then value of xx is _________

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Answer: 4

B⃗×C⃗=∣i^j^k^−143−8−13∣=15i^−21j^+33k^A⃗⋅(B⃗×C⃗)=(−xi^−6j^−2k^)⋅(15i^−21j^+33k^)0=−15x+126−6615x=60x=4\begin{aligned} & \vec{B} \times \vec{C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3\end{array}\right|=15 \hat{i}-21 \hat{j}+33 \hat{k} \\ & \vec{A} \cdot(\vec{B} \times \vec{C})=(-x \hat{i}-6 \hat{j}-2 \hat{k}) \cdot(15 \hat{i}-21 \hat{j}+33 \hat{k}) \\ & 0=-15 x+126-66 \\ & 15 x=60 \\ & x=4\end{aligned}

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