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

The position vector of the point of intersection of the medians (centroid) of a triangle,whose vertices are $A(1, 2, 3)$,$B(1, 0, 3)$,and $C(4, 1, -3)$ is

The points whose position vectors are $2\hat{i}+3\hat{j}+4\hat{k}$,$3\hat{i}+4\hat{j}+2\hat{k}$,and $4\hat{i}+2\hat{j}+3\hat{k}$ are the vertices of

$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

If $\vec{a}=(p, -2, 5)$ and $\vec{b}=(1, q, -3)$ are collinear vectors,then:

If the position vectors of three points are $a$,$b$,and $(3a - 2b)$,then these points are .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module