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Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise very short answer type question 7

Answers (1)

Answer:

 2 : 1

Hint:

 C = \left ( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n} \right )

Given:

 4x + 5y - 3z = 8  divides the line segment joining the points (-2, 1, 5) & (3, 3, 2)

Solution:

Let the coordinate of A (-2, 1, 5) and B(3, 3, 2) 

Let 4x + 5y - 3z = 8              …(i) divides the line segment AB at C in K : 1

C = internal section of AB

Co- ordinate of

C = \left ( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n} \right )

= \left ( \frac{3k-2}{k+1}, \frac{3k+1}{k+1}, \frac{2k+5}{k+1} \right )

Put the coordinate of C in (i)

\begin{aligned} &4 \left ( \frac{3k-2}{k+1} \right )+5 \left ( \frac{3k+1}{k+1} \right )-3 \left ( \frac{2k+5}{k+1} \right )=8\\ &\Rightarrow 12k -8+15k+5-6k-15=8k+8\\ &\Rightarrow 13k=26\\ &\Rightarrow k=2 \end{aligned}

Ratio K: 1 will be  2:1

Posted by

Gurleen Kaur

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