The point collinear with (1, -2, -3) and (2, | vedclass

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Question

(C) Let the given points be $A(1, -2, -3)$ and $B(2, 0, 0)$.The vector equation of the line passing through $A$ and $B$ is given by $\vec{r} = \vec{a}

Vedclass · Accepted Answer

$(0, -4, -6)$

Vedclass · Answer

(C) Let the given points be $A(1, -2, -3)$ and $B(2, 0, 0)$.The vector equation of the line passing through $A$ and $B$ is given by $\vec{r} = \vec{a} + t(\vec{b} - \vec{a})$,where $\vec{a} = \hat{i} - 2\hat{j} - 3\hat{k}$ and $\vec{b} = 2\hat{i}$.The direction vector is $\vec{v} = \vec{b} - \vec{a} = (2-1)\hat{i} + (0 - (-2))\hat{j} + (0 - (-3))\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.The line equation is $\vec{r} = (1\hat{i} - 2\hat{j} - 3\hat{k}) + t(\hat{i} + 2\hat{j} + 3\hat{k})$.This gives the parametric equations: $x = 1 + t$,$y = -2 + 2t$,$z = -3 + 3t$.For option $(C) (0, -4, -6)$,we set $x = 0 \implies 1 + t = 0 \implies t = -1$.Substituting $t = -1$ into $y$ and $z$: $y = -2 + 2(-1) = -4$ and $z = -3 + 3(-1) = -6$.Since the point $(0, -4, -6)$ satisfies the line equation,it is collinear.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

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