MHT CET · Physics · Mathematics in Physics
If \(\vec{A}=\hat{i}+\hat{j}+3 \hat{k}, \vec{B}=-\hat{i}+\hat{j}+4 \hat{k}\) and \(\vec{C}=2 \hat{i}-2 \hat{j}-8 \hat{k}\), then the angle between the vectors \(\vec{P}=\vec{A}+\vec{B}+\vec{C}\) and \(\vec{Q}=(\vec{A} \times \vec{B})\) is (in degree)
- A \(0^0\)
- B \(45^{\circ}\)
- C \(90^{\circ}\)
- D \(60^{\circ}\)
Answer & Solution
Correct Answer
(C) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
\(\vec{P}=\vec{A}+\vec{B}+\vec{C} = (\hat{i}+\hat{j}+3 \hat{k})+(-\hat{i}+\hat{j}+4 \hat{k})+(2 \hat{i}-2 \hat{j}-8 \hat{k}) = (1-1+2)\hat{i} + (1+1-2)\hat{j} + (3+4-8)\hat{k} = 2\hat{i}-\hat{k}\) \(\vec{Q}=\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ -1 & 1 & 4 \end{vmatrix} = \hat{i}((1)(4)-(3)(1))-\hat{j}((1)(4)-(3)(-1))+\hat{k}((1)(1)-(1)(-1)) = \hat{i}(4-3)-\hat{j}(4+3)+\hat{k}(1+1) = \hat{i}-7\hat{j}+2\hat{k}\)
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