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  • Provide solution for RD Sharma Maths Class 12 Chapter 26 Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 10.

Provide solution for RD Sharma Maths Class 12 Chapter 26 Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 10.

Answers (1)

Final Answer:

(a) \left [\frac{19}{8},\frac{57}{16},\frac{17}{16} \right ]

Hint: Bisection divides triangle in ratios

\frac{AB}{AC}=\frac{BD}{DC}

Given: A(3,2,0), B(5,3,2) and C(-9,6,-3) are vertices of triangle ABC

To find: Point D coordinates which meets BC at bisector of \angle BAC

Solution:  

  

As bisector AD meets BC at D, therefore

\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}

Using distance formula

\begin{aligned} &A B=\sqrt{(5-3)^{2}+(3-2)^{2}+(2-0)^{2}} \\ &=\sqrt{4+1+4} \\ &=\sqrt{9} \\ &=3 \end{aligned}

and,

\begin{aligned} &A C=\sqrt{(-9-3)^{2}+(6-2)^{2}+(-3-0)^{2}} \\ &=\sqrt{(-12)^{2}+4^{2}+3^{2}} \\ &=\sqrt{144+16+9} \\ &=\sqrt{169} \\ &=13 \end{aligned}

Therefore,

\frac{BD}{DC}=\frac{13}{8}

D divides BC in the ratio 3:13 internally

Using section formula coordinates of D are

\begin{aligned} &(x, y, z)=\left[\Rightarrow \frac{-9 \times 3+5 \times 13}{3+13}, \frac{6 \times 3+13 \times 3}{3+13}, \frac{-3 \times 3+13 \times 2}{3+13}\right] \\ &(x, y, z)=\left[\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right] \end{aligned}

Hence coordinate of D are

\left [\frac{19}{8},\frac{57}{16},\frac{17}{16} \right ]

Therefore option (a) is correct

Posted by

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