Find the angle between the pair of lines give | vedclass

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(A) The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$. Here,$\vec{b}_1 = \hat{i} + 2

Vedclass · Accepted Answer

$	heta=\cos ^{-1}\left(\frac{19}{21}ight)$

Vedclass · Answer

(A) The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$. Here,$\vec{b}_1 = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b}_2 = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$. The angle $	heta$ between the two lines is given by $\cos 	heta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} ight|$. First,calculate the dot product: $\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$. Next,calculate the magnitudes: $|\vec{b}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$. $|\vec{b}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$. Substituting these values into the formula: $\cos 	heta = \left| \frac{19}{3 	imes 7} ight| = \frac{19}{21}$. Therefore,$	heta = \cos^{-1}\left(\frac{19}{21}ight)$.

List-$I$	List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$	$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$	$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$	$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$	$IV$. $A, B, C$ are collinear

Find the angle between the pair of lines given by $\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ and $\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$.

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