The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors…

The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\overline{\mathrm{b}}=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ is one, then the value of $\lambda$ is
  1. 1
  2. -2
  3. -3
  4. 2

Solution

$\begin{aligned} & \overline{\mathrm{b}}+\overline{\mathrm{c}}=(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \text { Unit vector }=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{\overline{\mathrm{b}}+\overline{\mathrm{c}}}} \\ & =\frac{(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{(2+\lambda)^2+(-2)^2+2^2}} \\ & =\frac{(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{\lambda^2+4 \lambda+12}} \end{aligned}$ According to the given condition, $\begin{aligned} & (\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot\left(\frac{(2+\lambda) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{\lambda^2+4 \lambda+12}}\right)=1 \\ & \Rightarrow \frac{(2+\lambda)-4+2}{\sqrt{\lambda^2+4 \lambda+12}}=1 \\ & \Rightarrow \lambda=\sqrt{\lambda^2+4 \lambda+12} \\ & \Rightarrow \lambda^2=\lambda^2+4 \lambda+12 \\ & \Rightarrow 4 \lambda=-12 \\ & \Rightarrow \lambda=-3 \end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

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