What is the distance between the points (2, 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The distance $d$ between two points in polar coordinates $(r_1, 	heta_1)$ and $(r_2, 	heta_2)$ is given by the formula: $d = \sqrt{r_1^2 + r_2^2

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) The distance $d$ between two points in polar coordinates $(r_1, 	heta_1)$ and $(r_2, 	heta_2)$ is given by the formula: $d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos(	heta_1 - 	heta_2)}$ Given $r_1 = 2, 	heta_1 = 15^{\circ}$ and $r_2 = 1, 	heta_2 = 75^{\circ}$. Substituting the values: $d = \sqrt{2^2 + 1^2 - 2(2)(1) \cos(15^{\circ} - 75^{\circ})}$ $d = \sqrt{4 + 1 - 4 \cos(-60^{\circ})}$ Since $\cos(-60^{\circ}) = \cos(60^{\circ}) = 0.5$: $d = \sqrt{5 - 4(0.5)}$ $d = \sqrt{5 - 2}$ $d = \sqrt{3}$

What is the distance between the points $(2, 15^{\circ})$ and $(1, 75^{\circ})$ in polar coordinates?

Similar Questions

The ratio in which the straight line $3x + 4y = 6$ divides the join of the points $(2, -1)$ and $(1, 1)$ is

If the line $2x - y - 4 = 0$ divides the line segment joining the points $(2, -1)$ and $(1, -4)$ at the point $(a, b)$ in the ratio $m:n$,then $4(a - b(\frac{m}{n})^2) = $

The point which divides externally the line joining the points $(a + b, a - b)$ and $(a - b, a + b)$ in the ratio $a:b$ is

The point on the $y$-axis equidistant from the points $(3, 2)$ and $(-1, 3)$ is

If $A$ and $B$ are the points $(-3, 4)$ and $(2, 1)$,then the coordinates of the point $C$ on $AB$ produced such that $AC = 2 BC$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module