AP Inter 1st Year Maths Exercise 11d Solutions

Referring to the AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry Exercise 11d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11d

Question 1.
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.
Solution:
i) The coordinates of point R that divides the line segment joining the points P(x1, y1; z1) and Q(x2, y2, z2) internally in the ratio m : n are
\(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)\)
Let R(x, y, z) be a point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6) internally in the ratio 2:3.
Hence, x = \(\frac{2(1)+3(-2)}{2+3}\), y = \(\frac{2(-4)+3(3)}{2+3}\) and z = \(\frac{2(6)+3(5)}{2+3}\)
i.e x = \(\frac{-4}{5}\),y = \(\frac{1}{5}\) and z = \(\frac{27}{5}\)
Thus, the coordinates of the required point are \(\left(\frac{-4}{5}, \frac{1}{5}, \frac{27}{5}\right)\)

ii) The coordinates of point R that divides the line segment joining the points P(x1, y1, z1) and Q(x2, y2, z2) externally in the ratio m : n are
\(\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)\)
Let R(x, y, z) be a point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6)’externally in the ratio 2 : 3.
Hence, x = \(\frac{2(1)-3(-2)}{2-3}\), y = \(\frac{2(-4)-3(3)}{2-3}\) and z = \(\frac{2(6)-3(5)}{2-3}\)
i.e., x = -8, y = 17 and z = 3
Thus, the coordinates of the required point are (-8, 17, 3).

Question 2.
Given that P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear. Find the ratio in which 9 divides PR.
Solution:
Let point Q(5, 4, -6) divides the line segment joining points P(3, 2, -4) and R(9, 8, -10) in the ratio k : 1.
Therefore, by section formula,
(5, 4, -6) = \(\frac{1}{2}\)
Hence, \(\frac{9 k+3}{k+1}\) = 5 ⇒ 9k + 3 = 5k + 5
⇒ 4k = 2
⇒ k = \(\frac{2}{4}=\frac{1}{2}\)
Thus, point Q divides PR in the ratio 1:2.

AP Inter 1st Year Maths Exercise 11d Solutions

Question 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).
Solution:
Let the YZ plane divides the line segment joining points (-2, 4, 7) and (3, -5, 8) in the ratio k : 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{\mathrm{k}(3)-2}{\mathrm{k}+1}, \frac{\mathrm{k}(-5)+4}{\mathrm{k}+1}, \frac{\mathrm{k}(8)+7}{\mathrm{k}+1}\right)\)
On the YZ plane, the x-coordinate of any point is zero. 3k-2
\(\frac{3 k-2}{k+1}\) = 0
⇒ 3k – 2 = 0
⇒ k = \(\frac{2}{3}\)
Thus, the YZ-plane divides the line segment formed by joining the given points in the ratio 2 : 3.

Question 4.
Using section formula, show that the points A(2, -3, 4), B(-1, 2, 1) and are collinear.
Solution:
Let P be a point that divides AB in the ratio k : 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{k(-1)+2}{k+1}, \frac{k(2)-3}{k+1}, \frac{k(1)+4}{k+1}\right)\)
Now, we find the value of k at which point P coincides with point C.
By taking \(\\frac{-\mathrm{k}+2}{\mathrm{k}+1}\) = 0 , we obtain k = 2.
For k = 2, the coordinates of point P are (0, \(\frac{1}{3}\), 2)
i.e., (0, \(\frac{1}{3}\), 2) is a point that divides AB internally in the ratio 2:1. Hence, points A, B and C are collinear.

Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).
Solution:
Let A and B be the points that trisect the line segment joining the points P(4, 2, -6) and Q(10 -16, 6). Point A divides PQ in the ratio 1 : 2.
Therefore, by section formula, the coordinates of point A are given by
\(\left(\frac{1(10)+2(4)}{1+2}, \frac{1(-16)+2(2)}{1+2}, \frac{1(6)+2(-6)}{1+2}\right)\) = (6, -2, -2)
Point B divides PQ in the ratio 2:1.
Therefore, by section formula, the coordinates of point B are given by
\(\left(\frac{2(10)+1(4)}{1+2}, \frac{2(-16)+1(2)}{1+2}, \frac{2(6)+1(-6)}{1+2}\right)\) = (8, -10, 2)
Thus, (6, -4, -2) and (8, -10, 2) are the points that trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).