Question
If $\vec{\text{a}}$ be the position vector whose tip is (5, -3), find the coordinates of a point B such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$, the coordinates of A being (4, -1).

Answer

Let O be the origin and let P(5, -3) be the tip of the position vector $\vec{\text{a}}$. Then,
$\vec{\text{a}}=\overrightarrow{\text{OP}}=5\hat{\text{i}}-3\hat{\text{j}}$. Let the coordinate of B be (x, y) and A has coordinates (4, -1).
Therefore,
$\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A
$=\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}\big)-\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=(\text{x}-4)\hat{\text{i}}+(\text{y}+1)\hat{\text{j}}$
Now,
$\overrightarrow{\text{AB}}=\vec{\text{a}}$
$\Rightarrow(\text{x}-4)\hat{\text{i}}+(\text{y}+1)\hat{\text{j}}=5\hat{\text{i}}-3\hat{\text{j}}$
$\Rightarrow\text{x}-4=5\text{ and }\text{y}+1=-3$
$\Rightarrow\text{x}=9 \text{ and }\text{y}=-4$
Hence, the coordinates of B are (9, -4).

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

Solve the following differential equations:

$\text{x}\sqrt{1-\text{y}^2}\text{dx}+\text{y}\sqrt{1-\text{x}^2}\text{dy}=0$

Find the particular solution of the differential equation $\text{e}^\text{x}\tan\text{y dx}+(2-\text{e}^\text{x})\text{sec}^2\text{y dy}=0,$ given that $\text{y}=\frac{\pi}{4}\ \text{x} = 0.$
Show that $\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}=0,$ where $\alpha,\beta,\gamma$ are in A.P.
Discuss the continuity and differentiability of,

$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\cos\text{x}$
$\text{If y} = \sin (\log x) , \text{prove that}$

$x^{2} \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} + y =0$

Solve the following systems of linear equations by cramer's rule:
x - 4y - z = 11,
2x - 5y + 2z = 39,
-3x + 2y + z = 1
Show that the points A, B, C with position vectors $2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \hat{\text{i}} - 3\hat{\text{j}} - 5\hat{\text{k}} \text{ and } 3\hat{\text{i}} - 4\hat{\text{j}} - 4\hat{\text{k}}$ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
Evaluate the following definite integrals:
$\int\limits_{1}^{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
Evaluate:
$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)$