Question
Prove that the points having position vectors $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$ are collinear.
✓
Answer
Let A, B, C be the points with position vectors $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=-6\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
$=-3\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-3\overrightarrow{\text{AB}}$
So, $\overrightarrow{\text{AB}}$ 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, A, B, C are collinear.
Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=-6\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
$=-3\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-3\overrightarrow{\text{AB}}$
So, $\overrightarrow{\text{AB}}$ 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, A, B, 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.
Explore more
Similar questions
In Figure ABCD is a regular hexagon, which vectors are:
- Collinear.
- Equal.
- Co-initial.
- Collinear but not equal.
If $\text{xy}=1,$ prove that $\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
→Two numbers are selected at random from integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.
Using definite intergeals, find the area of the region bounded by the following curves, after making a rough sketch y = 1 + |x + 1|, x = -2, y = 0.
Evaluate the following integrals:$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
→Differentiate the following functions with respect to x:
$\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big), <\text{x}<\infty$
→$\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big), <\text{x}<\infty$
Differentiate $(\log\text{x})^\text{x}$ with respect to x.
→$\text{if x} = a \cos\theta + b \sin\theta, y = a\sin\theta - b\cos\theta, \text{show that y} ^{2} \frac{d^{2}y}{dx^{2}}- x \frac{dy}{dx} + y = 0.$
→Using properties of determinants, prove that:
$\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$
→$\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$
If a, b, c are the langths of sides, BC, CA and AB of a triangle ABC, prove that $\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}=\vec{\text{0}}$ and deduce that $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}.$
→