Find the angle between the following pairs of | vedclass

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

Question

(A) Let $Q$ be the angle between the given lines.The angle between the given lines is given by the formula,$\cos Q = \left| \frac{\vec{b}_{1} \cdot \v

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $Q$ be the angle between the given lines.The angle between the given lines is given by the formula,$\cos Q = \left| \frac{\vec{b}_{1} \cdot \vec{b}_{2}}{|\vec{b}_{1}| |\vec{b}_{2}|} ight|$.The given lines are parallel to the vectors $\vec{b}_{1} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and $\vec{b}_{2} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ respectively.Calculate the magnitudes of the vectors:$|\vec{b}_{1}| = \sqrt{3^{2} + 2^{2} + 6^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.$|\vec{b}_{2}| = \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Calculate the dot product of the vectors:$\vec{b}_{1} \cdot \vec{b}_{2} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k})$$= (3 	imes 1) + (2 	imes 2) + (6 	imes 2) = 3 + 4 + 12 = 19$.Substitute the values into the formula:$\cos Q = \frac{19}{7 	imes 3} = \frac{19}{21}$.Therefore,$Q = \cos ^{-1}\left(\frac{19}{21}ight)$.

Find the angle between the following pairs of lines:
$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$

Similar Questions

If $|a \times b| = 4$ and $|a \cdot b| = 2$,then $|a|^2 |b|^2 = $

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

In quadrilateral $ABCD$,$\overrightarrow{AB}=\vec{a}$,$\overrightarrow{BC}=\vec{b}$,$\overrightarrow{DA}=\vec{a}-\vec{b}$. $M$ is the midpoint of $BC$ and $X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5} \overrightarrow{DM}$. Then the points $A, X$ and $C$:

Let $a = i + 2j + k$,$b = i - j + k$,and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is:

Let $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}-2 \hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and the orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$,then find $|\vec{x}-\vec{y}|$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the angle between the following pairs of lines:$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$