Let A(4,3,5), B(1,-2,1), C(3,2,1) be the vert | vedclass

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

Question

(B) First,calculate the lengths of sides $AB$ and $AC$ using the distance formula:$AB = \sqrt{(4-1)^2 + (3-(-2))^2 + (5-1)^2} = \sqrt{3^2 + 5^2 + 4^2}

Vedclass · Accepted Answer

$\frac{3 \sqrt{5}}{4}$

Vedclass · Answer

(B) First,calculate the lengths of sides $AB$ and $AC$ using the distance formula:$AB = \sqrt{(4-1)^2 + (3-(-2))^2 + (5-1)^2} = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$$AC = \sqrt{(4-3)^2 + (3-2)^2 + (5-1)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$According to the Angle Bisector Theorem,the internal bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides containing the angle:$\frac{BD}{DC} = \frac{AB}{AC} = \frac{5\sqrt{2}}{3\sqrt{2}} = \frac{5}{3}$Thus,point $D$ divides $BC$ internally in the ratio $5:3$. Using the section formula,the coordinates of $D$ are:$D = \left( \frac{5(3) + 3(1)}{5+3}, \frac{5(2) + 3(-2)}{5+3}, \frac{5(1) + 3(1)}{5+3} \right) = \left( \frac{18}{8}, \frac{4}{8}, \frac{8}{8} \right) = \left( \frac{9}{4}, \frac{1}{2}, 1 \right)$Now,calculate the length $CD$ using the distance formula between $C(3,2,1)$ and $D(\frac{9}{4}, \frac{1}{2}, 1)$:$CD = \sqrt{(\frac{9}{4} - 3)^2 + (\frac{1}{2} - 2)^2 + (1 - 1)^2}$$CD = \sqrt{(-\frac{3}{4})^2 + (-\frac{3}{2})^2 + 0^2} = \sqrt{\frac{9}{16} + \frac{9}{4}} = \sqrt{\frac{9 + 36}{16}} = \sqrt{\frac{45}{16}} = \frac{3\sqrt{5}}{4}$

Let $A(4,3,5), B(1,-2,1), C(3,2,1)$ be the vertices of a triangle $ABC$. If the internal bisector of $\angle BAC$ meets the side $BC$ at $D$,then $CD=$

Similar Questions

Points $A(3, 2, 4)$,$B\left(\frac{33}{5}, \frac{28}{5}, \frac{38}{5}\right)$ and $C(9, 8, 10)$ are given. The ratio in which $B$ divides $\overline{AC}$ is

If $A(4,3,2), B(5,4,6), C(-1,-1,5)$ are the vertices of a triangle,then the coordinates of the point in which the bisector of the angle $A$ meets the side $BC$ is

The point dividing the line segment joining $(3, -2, 1)$ and $(-2, 3, 11)$ in the ratio $2:3$ is

If $P$ divides the line segment joining the points $A(1, 2, -1)$ and $B(-1, 0, 1)$ externally in the ratio $1: 2$ and $Q = (1, 3, -1)$,then $PQ =$

If $A(4,7,8)$,$B(2,3,4)$,and $C(2,5,7)$ are the vertices of $\triangle ABC$,then the length of the internal bisector of the angle $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module