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3-1.Vectors
hard
ત્રણ સદિશો $\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k}), \vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})$ અને $\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k})$ માટે જો $\vec{A} \cdot(\vec{B} \times \vec{C})=0$ હોય તો $x$ નું મૂલ્ચ. . . . . .છે.
A
$2$
B
$4$
C
$6$
D
$8$
(JEE MAIN-2024)
Solution
$\vec{B} \times \vec{C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3\end{array}\right|=15 \hat{i}-21 \hat{j}+33 \hat{k}$
$\vec{A} \cdot(\vec{B} \times \vec{C})=(-x \hat{i}-6 \hat{j}-2 \hat{k}) \cdot(15 \hat{i}-21 \hat{j}+33 \hat{k})$
$0=-15 x+126-66$
$15 x=60$
$x=4$
Standard 11
Physics
