Lines vec{r} = vec{a} + lambdavec{b} and vec{ | vedclass

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(A) Step $1$: Since the lines intersect at $(2, 4, -4)$, the point must satisfy both equations.Step $2$: For the first line, $\vec{a} + \lambda\vec{b}

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$5$

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(A) Step $1$: Since the lines intersect at $(2, 4, -4)$, the point must satisfy both equations.Step $2$: For the first line, $\vec{a} + \lambda\vec{b} = (2, 4, -4)$. For the second line, $\vec{b} + \mu\vec{a} = (2, 4, -4)$.Step $3$: Subtracting the two equations: $(\vec{a} - \vec{b}) + (\lambda\vec{b} - \mu\vec{a}) = 0$. This implies $\vec{a}(1 - \mu) = \vec{b}(1 - \lambda)$.Step $4$: Since the lines intersect at the same point, $\vec{a} + \lambda\vec{b} = \vec{b} + \mu\vec{a} \implies \vec{a}(1 - \mu) = \vec{b}(1 - \lambda)$. If $\vec{a}$ and $\vec{b}$ are not collinear, then $1-\mu = 0$ and $1-\lambda = 0$, so $\lambda = 1$ and $\mu = 1$.Step $5$: Substituting $\lambda = 1$ into the first equation: $\vec{a} + \vec{b} = (2, 4, -4)$.Step $6$: We are given $|\vec{a} - \vec{b}| = 4$. Squaring both sides: $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 16$.Step $7$: From $\vec{a} + \vec{b} = (2, 4, -4)$, we have $|\vec{a} + \vec{b}|^2 = 2^2 + 4^2 + (-4)^2 = 4 + 16 + 16 = 36$. So $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 36$.Step $8$: Subtracting the two equations: $(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}) - (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) = 36 - 16 \implies 4\vec{a} \cdot \vec{b} = 20 \implies \vec{a} \cdot \vec{b} = 5$.

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

Lines $\vec{r} = \vec{a} + \lambda\vec{b}$ and $\vec{r} = \vec{b} + \mu\vec{a}$ intersect at point $(2, 4, -4)$. If $|\vec{a} - \vec{b}| = 4$, then $\vec{a} \cdot \vec{b} =$

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