Q72 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
The vector $\bar{r}$ whose magnitude is $3\sqrt{2}$ units and which makes angles of $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$ with the positive y- and z-axes respectively is....
- A$\hat{i} \pm 3\hat{j}$
- B$\hat{i} \pm \hat{j}$
- C$-\hat{i} \pm \hat{j}$
- Correct answerD$\pm 3\hat{i} + 3\hat{j}$
Solution
Let the direction cosines of the vector $\bar{r}$ be $l, m, n$.
Given that the vector makes angles $\beta = \dfrac{\pi}{4}$ and $\gamma = \dfrac{\pi}{2}$ with the positive y- and z-axes respectively.
$m = \cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
$n = \cos\left(\dfrac{\pi}{2}\right) = 0$
Using the relation $l^2 + m^2 + n^2 = 1$:
$l^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2 + 0 = 1$
$l^2 + \dfrac{1}{2} = 1 \Rightarrow l = \pm\dfrac{1}{\sqrt{2}}$
The vector $\bar{r}$ is given by $|\bar{r}|(l\hat{i} + m\hat{j} + n\hat{k})$.
Substituting the values:
$\bar{r} = 3\sqrt{2} \left( \pm\dfrac{1}{\sqrt{2}}\hat{i} + \dfrac{1}{\sqrt{2}}\hat{j} + 0\hat{k} \right)$
$\bar{r} = \pm 3\hat{i} + 3\hat{j}$
Answer: $\pm 3\hat{i} + 3\hat{j}$
Given that the vector makes angles $\beta = \dfrac{\pi}{4}$ and $\gamma = \dfrac{\pi}{2}$ with the positive y- and z-axes respectively.
$m = \cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
$n = \cos\left(\dfrac{\pi}{2}\right) = 0$
Using the relation $l^2 + m^2 + n^2 = 1$:
$l^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2 + 0 = 1$
$l^2 + \dfrac{1}{2} = 1 \Rightarrow l = \pm\dfrac{1}{\sqrt{2}}$
The vector $\bar{r}$ is given by $|\bar{r}|(l\hat{i} + m\hat{j} + n\hat{k})$.
Substituting the values:
$\bar{r} = 3\sqrt{2} \left( \pm\dfrac{1}{\sqrt{2}}\hat{i} + \dfrac{1}{\sqrt{2}}\hat{j} + 0\hat{k} \right)$
$\bar{r} = \pm 3\hat{i} + 3\hat{j}$
Answer: $\pm 3\hat{i} + 3\hat{j}$
Hence, the correct answer is option: 'D'.