Q80 / 180 · COMEDK 2022 · Mathematics
If $\mathrm{p}=\hat{i}+\hat{j}, \mathrm{q}=4 \hat{k}-\hat{j}$ and $\mathrm{r}=\hat{i}+\hat{k}$, then the unit vector in the direction of $3 \mathrm{p}+\mathrm{q}-2 \mathrm{r}$ is
- Correct answerA$\dfrac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})$
- B$\dfrac{1}{3}(\hat{i}-2 \hat{j}-2 \hat{k})$
- C$\dfrac{1}{3}(\hat{i}-2 \hat{j}+2 \hat{k})$
- D$\hat{i}+2 \hat{j}+2 \hat{k}$
Solution
Given vectors are $\vec{p} = \hat{i} + \hat{j}$, $\vec{q} = -\hat{j} + 4\hat{k}$, and $\vec{r} = \hat{i} + \hat{k}$.
Calculate the vector $\vec{v} = 3\vec{p} + \vec{q} - 2\vec{r}$:
$\vec{v} = 3(\hat{i} + \hat{j}) + (-\hat{j} + 4\hat{k}) - 2(\hat{i} + \hat{k})$
$\vec{v} = (3\hat{i} + 3\hat{j}) + (-\hat{j} + 4\hat{k}) - (2\hat{i} + 2\hat{k})$
$\vec{v} = (3 - 2)\hat{i} + (3 - 1)\hat{j} + (4 - 2)\hat{k}$
$\vec{v} = \hat{i} + 2\hat{j} + 2\hat{k}$
The magnitude of vector $\vec{v}$ is $|\vec{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The unit vector in the direction of $\vec{v}$ is $\hat{u} = \dfrac{\vec{v}}{|\vec{v}|} = \dfrac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$.
Answer: $\dfrac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})$
Calculate the vector $\vec{v} = 3\vec{p} + \vec{q} - 2\vec{r}$:
$\vec{v} = 3(\hat{i} + \hat{j}) + (-\hat{j} + 4\hat{k}) - 2(\hat{i} + \hat{k})$
$\vec{v} = (3\hat{i} + 3\hat{j}) + (-\hat{j} + 4\hat{k}) - (2\hat{i} + 2\hat{k})$
$\vec{v} = (3 - 2)\hat{i} + (3 - 1)\hat{j} + (4 - 2)\hat{k}$
$\vec{v} = \hat{i} + 2\hat{j} + 2\hat{k}$
The magnitude of vector $\vec{v}$ is $|\vec{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The unit vector in the direction of $\vec{v}$ is $\hat{u} = \dfrac{\vec{v}}{|\vec{v}|} = \dfrac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$.
Answer: $\dfrac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})$
Hence, the correct answer is option: 'A'.