Question
Show that the points A, B, C with position vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ and $-7\vec{\text{b}}+10\vec{\text{c}}$ are collinear.

Answer

We have, A, B, C with position vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ and $-7\vec{\text{b}}+10\vec{\text{c}}$ Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}-\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$
$=\vec{\text{a}}+5\vec{\text{b}}-7\vec{\text{c}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-7\vec{\text{b}}+10\vec{\text{c}}-2\vec{\text{a}}-3\vec{\text{b}}+4\vec{\text{c}}$
$=-2\vec{\text{a}}-10\vec{\text{b}}+14\vec{\text{c}}$
$=-2\big(\vec{\text{a}}+5\vec{\text{b}}-7\vec{\text{c}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-2\overrightarrow{\text{AB}}$
Hence, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors.
But B is a point common to them.
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are collinear.
Hence, points A, B and C are collinear.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Evaluate the following integrals:
$\int\limits_{0}^{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$
Bag $I$ contains $3$ black and $2$ white balls, Bag $II$ contains $2$ black and $4$ white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
Solve the following differential equation $\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$
Evaluate the following integrals:
$\int\frac{\text{x}}{\text{x}^4+2\text{x}^2+3}\text{dx}$
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD, BC and AD. Show that $\overrightarrow{\text{PA}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PC}}+\overrightarrow{\text{PD}}=4\ \overrightarrow{\text{PQ}}$, where P ia any point.
Find the equation of the plane through the line of intersection of the planes $x + 2y + 3z + 4 = 0$ and $x - y + z + 3 = 0$ and passing through the origin.
The random variable $X$ can take only the values $0, 1, 2.$ Given that $P(X = 0) = P (X = 1) = p$ and that $E(X^2) = E[X], $ find the value of $p.$
Without expanding, prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$\text{A}=\begin{bmatrix}3&-4&2\\ 2&3&5\\ 1&0&1\end{bmatrix}$, find $A^{-1}$ and hence solve the following system of equations:
$3x - 4y +2z = -1, 2x + 3y + 5z = 7, x + z = 2$