Ncert Solutions Maths class 11th Maths Class 11th Introduction to Three Dimensional Geometry

Class 11 Math Ex 12.3 NCERT Solution

10. Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.

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4 months ago

10. i. Let P(x, y, z) be the point which divides line segment joining (–2, 3, 5) and (1, –4, 6) internally in the ratio 2 : 3

Therefore,

x = $\frac{2 (1) + 3 (- 2)}{2 + 3}$ = $\frac{2 - 6}{5}$ = $\frac{- 4}{5}$

y = $\frac{2 (- 4) + 3 (3)}{2 + 3}$ = $\frac{- 8 + 9}{5}$ = $\frac{1}{5}$

z = $\frac{2 (6) + 3 (5)}{2 + 3}$ = $\frac{12 + 15}{5}$ = $\frac{27}{5}$

Thus, the required points are $(\frac{- 4}{5}, \frac{1}{5}, \frac{27}{5})$

ii. Let P(x, y, z) be the point which divides line segment joining (–2, 3, 5) and (1, –4, 6) externally in the ratio 2 : 3

Therefore,

x = $\frac{2 (1) - 3 (- 2)}{2 - 3}$ = $\frac{2 + 6}{- 1}$ = –8

y = $\frac{2 (- 4) - 3 (3)}{2 - 3}$ = $\frac{- 8 - 9}{- 1}$ = 17

z = $\frac{2 (6) - 3 (5)}{2 - 3}$ = $\frac{12 - 15}{- 1}$ = 3

Thus, the required points are (–8, 17, 3).

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Any point on line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2} = r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

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$P (4, 6, 7) A (1, 1, 9) d i s t A P = \sqrt[]{9 + 25 + 4} = \sqrt[]{38}$

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Class 11 Math Ex 12.3 NCERT Solution

10. Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.