The angle between the lines bar{r}=(3 hat{i}+ | vedclass

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

Question

(D) The direction vectors of the two lines are $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.The angle $	h

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{19}{21}\right)$

Vedclass · Answer

(D) The direction vectors of the two lines are $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.The angle $	heta$ between the lines is given by the formula $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.Next,calculate the magnitudes: $|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.$|\vec{b}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.Therefore,$\cos 	heta = \frac{19}{3 	imes 7} = \frac{19}{21}$.Thus,$	heta = \cos ^{-1}\left(\frac{19}{21}ight)$.

The angle between the lines $\bar{r}=(3 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ and $\bar{r}=(5 \hat{i}-2 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$ is:

Similar Questions

The equation of the line passing through the points $(3, 2, 4)$ and $(4, 5, 2)$ is

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

The angle between the straight lines $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4}$ and $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{-3}$ is ......... $^o$

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

If the points $A(-1, 3, 2)$,$B(-4, 2, -2)$,and $C(5, 5, \lambda)$ are collinear,then $\lambda = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module