The angle made by the position vector of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The position vector of the point $P(5, -4, -3)$ is $\vec{r} = 5\hat{i} - 4\hat{j} - 3\hat{k}$.The direction cosines of a vector $\vec{r} = a\hat{i

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) The position vector of the point $P(5, -4, -3)$ is $\vec{r} = 5\hat{i} - 4\hat{j} - 3\hat{k}$.The direction cosines of a vector $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ are given by $\cos \alpha = \frac{a}{|\vec{r}|}$,$\cos \beta = \frac{b}{|\vec{r}|}$,and $\cos \gamma = \frac{c}{|\vec{r}|}$,where $\alpha, \beta, \gamma$ are the angles made with the $X, Y, Z$ axes respectively.First,calculate the magnitude of the vector $\vec{r}$:$|\vec{r}| = \sqrt{5^2 + (-4)^2 + (-3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.The angle $\alpha$ with the positive $X$-axis is given by:$\cos \alpha = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

The angle made by the position vector of the point $(5, -4, -3)$ with the positive direction of the $X$-axis is:

Similar Questions

In the figure,a vector $x$ satisfies the equation $x - w = v$. Then $x =$

If $\overline{a} = m \overline{b} + n \overline{c}$,where $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$,then $m + n =$

In a parallelogram $ABCD$,if the diagonals are $\overline{AC}$ and $\overline{BD}$,then $\overline{AC} - \overline{BD} = \dots$

The direction cosines of the resultant of the vectors $(i + j + k)$,$(-i + j + k)$,$(i - j + k)$ and $(i + j - k)$ are

The vectors $a$,$b$,and $a + b$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module