Question
Using vector method, prove that the point is collinear:
A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7)

Answer

Given the points A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$=4\hat{\text{i}}+4\hat{\text{j}}+8\hat{\text{k}}$
$=2\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given 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

The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k} \text{ and } 2\hat{i} + 2\hat{j} + 3\hat{k}.$ Find the two unit vectors parallel its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Solve the following differential equations:$(\text{y + xy})\text{dx}+(\text{x}-\text{xy}^2)\text{dy}=0$
There are three categories of students in a class of 60 students:
A : Very hardworking
B : Regular but not so hardworking
C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C.
It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
If $\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$ are two square matrices, find AB and hence solve the system of linear equations:
x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$\text{f}(\text{x})=(\text{x}-2)\sqrt{\text{x}-1}\ \text{in}\ [1,9]$
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
If $\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta$ Show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$
Evaluate the following integrals:$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
If $\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that