Q128 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Physics

Find the angle between the vectors $\mathbf{A}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{B}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.

  • A
    $0^{\circ}$
  • B
    $45^{\circ}$
  • C
    $60^{\circ}$
  • Correct answer
    D
    $90^{\circ}$

Solution

Given, $\mathbf{A}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{B}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ Let $\theta$ be the angle between $\mathbf{A}$ and $\mathbf{B}$. By using dot product, $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta \text {, }$ where, $A$ and $B$ are magnitudes of $\mathbf{A}$ and $\mathbf{B}$ respectively. Now, $\begin{aligned} & A=\sqrt{2^2+4^2+4^2}=6 \\ & B=\sqrt{4^2+2^2+(-4)^2}=6 \end{aligned}$ Substituting the values, in dot product, we get $\begin{aligned} & (2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=(6)(6) \cos \theta \\ & 8+8-16=36 \cos \theta \\ & \Rightarrow \quad \cos \theta=0 \\ & \Rightarrow \quad \theta=90^{\circ} \end{aligned}$

Hence, the correct answer is option: 'D'.

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