Let $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$ be two vectors, and…

Let $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$ be two vectors, and $\overrightarrow{\mathrm{r}}$ be a vector along the vector $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$ such that $|\overrightarrow{\mathrm{r}}|=\sqrt{74}$. If the direction of $\vec{r}$ is opposite to that of $3 \vec{a}-2 \vec{b}$, then $\overrightarrow{\mathrm{r}}=$
  1. $-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
  2. $4 \hat{i}+7 \hat{j}-3 \hat{k}$
  3. $-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
  4. $4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Solution

Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$ Now, $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}=3(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})-2(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$ $=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$ Since $\overrightarrow{\mathrm{r}}$ along the vector $3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}$ So, $\vec{r}=x(3 \vec{a}-2 \vec{b})=x(4 \hat{i}-3 \hat{j}+7 \hat{k})$ Now, $|\vec{r}|=\sqrt{74} \Rightarrow \sqrt{16 x^2+9 x^2+49 x^2}=\sqrt{74}$ $\Rightarrow \sqrt{74} x^2=\sqrt{74} \Rightarrow x= \pm 1$ since, $\vec{r}$ is the opposite direction of $3 \vec{a}-2 \vec{b}$ So $\mathrm{x}=-1$ $\Rightarrow \overrightarrow{\mathrm{r}}=-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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