Q.The position vectors $\vec a,\vec b,\vec c$ of three points satisfy the relation $2\vec a-7\vec b+5\vec c=\vec 0$. Are these points collinear?

The position vector of A is $4\hat i+5\hat j$.

$\overrightarrow{BA}=\vec A-\vec B$, so $\vec A=\overrightarrow{BA}+\vec B$.

Given $\overrightarrow{BA}=3\hat i+2\hat j+\hat k$ and $\vec B=\hat i+3\hat j-\hat k$: …

The third vertex is $3G-(v_1+v_2) = -2\hat i-\hat j+9\hat k$.

For a triangle with vertices $v_1,v_2,v_3$, the centroid is $G=\dfrac{v_1+v_2+v_3}{3}$, so $v_3 = 3G-v_1-v_2$.

Here $v_1=3\hat i+4\hat j-4\hat k$, $v_2=2\hat i+3\hat j+4\hat k$, $G=\hat i+2\hat j+3\hat k$.

$3G = 3\hat i+6\hat j+9\hat k$.

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Equating the slope of the first two points to the slope of the last two points gives $a=8$.

The points, as coordinates from their position vectors, are $P_1(10,3)$, $P_2(12,-5)$, $P_3(a,11)$.

Slope of $P_1P_2 = \dfrac{-5-3}{12-10} = \dfrac{-8}{2} = -4$.

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Any point on the line through A and B has position vector $(1-t)\vec a+t\vec b$, whose coefficients always add to 1; checking each option, only $\dfrac{2\vec a+\vec b}{3}$ has coefficients summing to $1$.

The line through points A (position vector $\vec a$) and B (position vector $\vec b$) is parametrized as $\vec r = \vec a + t(\vec b-\vec a) = (1-t)\vec a + t\vec b$ for $t\in\mathbb{R}$. A necessary condition for a vector to be a point on this line is that its $\vec a$- and $\vec b$-coefficients sum to exactly 1.

Check (a) $\dfrac{2\vec a+\vec b}{3} = \dfrac{2}{3}\vec a + \dfrac{1}{3}\vec b$: coefficients sum to $\dfrac{2}{3}+\dfrac{1}{3}=1$. ✓ (this is the point dividing AB in ratio $1:2$ from A).

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