Vectors | Physics JEE previous year questions with Solutions

Question

For three vectors $\vec{A} = (- x \hat{i} - 6 \hat{j} - 2 \hat{k}), \vec{B} = (- \hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\vec{C} = (- 8 \hat{i} - \hat{j} + 3 \hat{k})$ , if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ then value of $x$ is _______ . [2024]

Smart Achievers · Accepted Answer

(4) $\vec{B} = - \hat{i} + 4 \hat{j} + 3 \hat{k} a n d \vec{C} = - 8 \hat{i} - \hat{j} + 3 \hat{k}$

$\vec{C} = (- 8 \hat{i} - \hat{j} + 3 \hat{k})$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 4 & 3 \\ - 8 & - 1 & 3 \end{matrix} |$ $= 15 \hat{i} - 21 \hat{j} + 33 \hat{k}$

$\vec{A} \cdot (\vec{B} \times \vec{C}) = (- x \hat{i} - 6 \hat{j} - 2 \hat{k}) \cdot (15 \hat{i} - 21 \hat{j} + 33 \hat{k})$

$0 = - 15 x + 126 - 66$

$\Rightarrow 15 x = 60 \Rightarrow x = 4$

Solution

Q.
For three vectors $\vec{A} = (- x \hat{i} - 6 \hat{j} - 2 \hat{k}), \vec{B} = (- \hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\vec{C} = (- 8 \hat{i} - \hat{j} + 3 \hat{k})$ , if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ then value of $x$ is _______ . [2024]

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Solution

Q. For three vectors A→=(-xi^-6j^-2k^), B→=(-i^+4j^+3k^) and C→=(-8i^-j^+3k^), if A→·(B→×C→)=0 then value of x is _______ . [2024]

Ans. (4) B→=-i^+4j^+3k^ and C→=-8i^-j^+3k^ B→×C→=|i^j^k^-143-8-13|=15i^-21j^+33k^ A→·(B→×C→)=(-xi^-6j^-2k^)·(15i^-21j^+33k^) 0=-15x+126-66 ⇒15x=60⇒x=4

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Q.
For three vectors $\vec{A} = (- x \hat{i} - 6 \hat{j} - 2 \hat{k}), \vec{B} = (- \hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\vec{C} = (- 8 \hat{i} - \hat{j} + 3 \hat{k})$ , if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ then value of $x$ is _______ . [2024]