The vector of magnitude 6 units and perpendicular to vectors $2 \hat{i}+\hat{j}-3 \hat{k}$ and $\hat{i}-2…

The vector of magnitude 6 units and perpendicular to vectors $2 \hat{i}+\hat{j}-3 \hat{k}$ and $\hat{i}-2 \hat{j}+\hat{k}$ is
  1. $\quad 2 \sqrt{3}(-\hat{i}+\hat{j}+\hat{k})$
  2. $\quad 2 \sqrt{3}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$
  3. $2 \sqrt{3}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
  4. $\quad 2 \sqrt{3}(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

Solution

Let the required vector be $\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$. Then, $|\hat{r}|=6$ $\Rightarrow x^2+y^2+z^2=36$
Now, $\bar{r}$ is perpendicular to vectors $\bar{a}=2 \hat{i}+\hat{j}-3 \hat{k}$ and $\bar{b}=\hat{i}-2 \hat{j}+\hat{k}$. $\begin{array}{ll} \therefore & \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -3 \\ 1 & -2 & 1 \end{array}\right|=-5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ \therefore & x=y=\mathrm{z} \\ \therefore & \text { Let } x=y=\mathrm{z}=\lambda \end{array}$ Let $x=y=z=\lambda$. From (i), we get $\begin{array}{ll} & 3 \lambda^2=36 \\ \therefore \quad & \lambda=2 \sqrt{3} \end{array}$ $\therefore \quad$ Required vector is $2 \sqrt{3}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$

Asked in: MHT CET 2024 (10 May Shift 2)

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