Q73 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Mathematics

Let $\mathbf{F}=2 \hat{i}+2 \hat{j}+5 \hat{k}, A=(1,2,5)$, $B=(-1,-2,-3)$ and $\mathbf{B A} \times \mathbf{F}=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$, then $\lambda=$

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • Correct answer
    D
    $-2$

Solution

Given, $\begin{aligned} & \mathbf{F}=2 i+2 j+5 k, A=(1,2,5) \\ & B=(-1,-2,-3) \\ & \mathbf{B A}=\mathbf{A}-\mathbf{B}=(\hat{i}+2 \hat{j}+5 \hat{k})-(-\hat{i}-2 \hat{j}-3 \hat{k}) \\ & \mathbf{B A}=2 \hat{i}+4 \hat{j}+8 \hat{k}\end{aligned}$ $\mathbf{B A} \times \mathbf{F}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 8 \\ 2 & 2 & 5\end{array}\right|$ $\begin{aligned} & =\hat{i}(20-16)-\hat{j}(10-16)+\hat{k}(4-8) \\ & =4 \hat{i}+6 \hat{j}-4 \hat{k}\end{aligned}$ So, $4 \hat{i}+6 \hat{j}-4 \hat{k}=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$ $\because \quad 2 \lambda=-4$ $\lambda=-2$

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

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