GSEB Class 11 Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Misc. Questions

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
Answer:
(i) Let P(\(x, y, z\)) be the point that divides the line segment joining the points (- 2, 3, 5) and (1, - 4, 6) with an internal ratio of 2 : 3. The coordinates for point P are calculated using the ratio \(m : n\).
\[ x = \frac{mx_2 + nx_1}{m+n}, y = \frac{my_2 + ny_1}{m+n}, z = \frac{mz_2 + nz_1}{m+n} \]
So, we have:
\( x = \frac{2 \times 1 + 3 \times (-2)}{2+3} = \frac{2-6}{5} = \frac{-4}{5} \)
\( y = \frac{2 \times (-4) + 3 \times 3}{2+3} = \frac{-8+9}{5} = \frac{1}{5} \)
\( z = \frac{2 \times 6 + 3 \times 5}{2+3} = \frac{12+15}{5} = \frac{27}{5} \)
Therefore, point P is \( (\frac{-4}{5}, \frac{1}{5}, \frac{27}{5}) \).
(ii) If point P divides AB externally in the ratio \(m : n\), then the coordinates for P are:
\[ P = (\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n}) \]
Calculating the values:
\( x = \frac{2 \times 1 - 3 \times (-2)}{2-3} = \frac{2+6}{-1} = -8 \)
\( y = \frac{2 \times (-4) - 3 \times 3}{2-3} = \frac{-8-9}{-1} = 17 \)
\( z = \frac{2 \times 6 - 3 \times 5}{2-3} = \frac{12-15}{-1} = 3 \)
The point (- 8, 17, 3) divides AB externally in the ratio 2 : 3.
In simple words: To find the point that divides a line segment, you use a special formula. For internal division, you add the terms, and for external division, you subtract them. Just plug in the given coordinates and ratios to get the new point's location.

Exam Tip: Remember the difference between the internal and external division formulas. A common mistake is using the wrong sign in the denominator or numerator for external division.

Question 2. Given that P(3, 2, - 4), Q(5, 4, - 6) and R(9, 8, - 10) are collinear. Find the ratijo in which Q divides PR.
Answer: Let Q divide PR in the ratio \(k : 1\). We can use the section formula to determine the value of \(k\) by comparing the coordinates.
For the x-coordinate of Q:
\( 5 = \frac{k \times 9 + 1 \times 3}{k+1} \)
\( 5(k+1) = 9k+3 \)
\( 5k+5 = 9k+3 \)
\( 4k = 2 \)
\( k = \frac{2}{4} = \frac{1}{2} \)
For the y-coordinate of Q:
\( 4 = \frac{k \times 8 + 1 \times 2}{k+1} \)
\( 4(k+1) = 8k+2 \)
\( 4k+4 = 8k+2 \)
\( 4k = 2 \)
\( k = \frac{2}{4} = \frac{1}{2} \)
For the z-coordinate of Q:
\( -6 = \frac{k \times (-10) + 1 \times (-4)}{k+1} \)
\( -6(k+1) = -10k-4 \)
\( -6k-6 = -10k-4 \)
\( 4k = 2 \)
\( k = \frac{2}{4} = \frac{1}{2} \)
Since the value of \(k\) is the same for all three coordinates, it shows that point Q lies on PR and divides it in the ratio 1 : 2.
In simple words: When three points are in a straight line, one point splits the segment formed by the other two. We use a formula to figure out this splitting ratio. By comparing the x, y, and z numbers, we found that Q divides PR in a 1:2 ratio.

Exam Tip: To show that a point divides a segment in a certain ratio, always check that the ratio derived from the x, y, and z coordinates is consistent. If they are all the same, the point divides the segment in that ratio.

Question 3. Find the ratio in which YZ-plane divides the line segment joining the points (- 2, 4, 7) and (3, - 5, 8).
Answer: Let the line segment connecting points A(- 2, 4, 7) and B(3, - 5, 8) be split by the YZ-plane at point P(\(0, y_1, z_1\)) in the ratio \(k : 1\).
Since the point lies on the YZ-plane, its x-coordinate must be 0.
Using the section formula for the x-coordinate:
\( 0 = \frac{k \times 3 + 1 \times (-2)}{k+1} \)
\( 0 = \frac{3k-2}{k+1} \)
This implies:
\( 3k-2 = 0 \)
\( 3k = 2 \)
\( k = \frac{2}{3} \)
Therefore, the YZ-plane divides the line segment AB in the ratio 2 : 3.
In simple words: The YZ-plane cuts through lines where the x-value is zero. We use this fact and a special formula to find the specific ratio in which this plane divides the line segment joining the two given points.

Exam Tip: When a plane divides a line segment, the coordinate corresponding to that plane is zero. For the YZ-plane, the x-coordinate is 0; for the XZ-plane, the y-coordinate is 0; and for the XY-plane, the z-coordinate is 0.

Question 4. Using section formula, show that the points A(2, - 3, 4), B(- 1, 2, 1) and C(0, \( \frac{1}{3} \), 2) are collinear.
Answer: Let's assume point C divides the line segment AB in the ratio \(k : 1\). We will use the section formula and see if a consistent \(k\) value can be found for all coordinates.
For the x-coordinate of C:
\( 0 = \frac{k \times (-1) + 1 \times 2}{k+1} \)
\( 0 = \frac{-k+2}{k+1} \)
\( -k+2 = 0 \)
\( k = 2 \)
Now, we check if this ratio holds for the y-coordinate of C:
\( \frac{1}{3} = \frac{k \times 2 + 1 \times (-3)}{k+1} \)
Substitute \(k=2\):
\( \frac{1}{3} = \frac{2 \times 2 - 3}{2+1} = \frac{4-3}{3} = \frac{1}{3} \)
The value matches. Next, we check for the z-coordinate of C:
\( 2 = \frac{k \times 1 + 1 \times 4}{k+1} \)
Substitute \(k=2\):
\( 2 = \frac{2 \times 1 + 4}{2+1} = \frac{2+4}{3} = \frac{6}{3} = 2 \)
The value matches here as well. Since the same ratio \(k=2\) satisfies all three coordinates, it means point C lies on the line segment AB and divides it in the ratio 2 : 1. Hence, points A, C, and B are collinear.
In simple words: To show that three points are on the same line, we check if one point divides the segment formed by the other two in a fixed ratio. If this ratio is the same for all x, y, and z coordinates, then the points are on one line.

Exam Tip: When proving collinearity using the section formula, it's crucial that the calculated ratio \(k\) is identical for all three coordinates (x, y, and z). If any of them differ, the points are not collinear.

Question 5. Find the coordinates of the points which trisect the line segment PQ formed by joining the points P(4, 2, - 6) and Q(10, - 16, 6).
Answer: To trisect a line segment means to divide it into three equal parts. So, there will be two points, A and B, that do this.
Let A(\(x_1, y_1, z_1\)) and B(\(x_2, y_2, z_2\)) be the points that trisect the line segment PQ. This implies that point A divides PQ in the ratio 1:2, and point B divides PQ in the ratio 2:1.
First, find the coordinates of point A (dividing PQ in 1:2 internally):
\( A = (\frac{1 \times 10 + 2 \times 4}{1+2}, \frac{1 \times (-16) + 2 \times 2}{1+2}, \frac{1 \times 6 + 2 \times (-6)}{1+2}) \)
\( A = (\frac{10+8}{3}, \frac{-16+4}{3}, \frac{6-12}{3}) \)
\( A = (\frac{18}{3}, \frac{-12}{3}, \frac{-6}{3}) \)
So, the coordinates of point A are \( (6, -4, -2) \).
Next, find the coordinates of point B (dividing PQ in 2:1 internally):
\( B = (\frac{2 \times 10 + 1 \times 4}{2+1}, \frac{2 \times (-16) + 1 \times 2}{2+1}, \frac{2 \times 6 + 1 \times (-6)}{2+1}) \)
\( B = (\frac{20+4}{3}, \frac{-32+2}{3}, \frac{12-6}{3}) \)
\( B = (\frac{24}{3}, \frac{-30}{3}, \frac{6}{3}) \)
So, the coordinates of point B are \( (8, -10, 2) \).
Thus, the points that trisect the line segment PQ are A(6, -4, -2) and B(8, -10, 2).
In simple words: To split a line into three equal parts, you need two points. The first point is found by using a 1:2 ratio, and the second point uses a 2:1 ratio. You use the section formula to calculate the exact x, y, and z values for both of these points.

Exam Tip: When a line segment is trisected, remember that the two points divide the segment in ratios 1:2 and 2:1. Accurately applying the section formula for each ratio is key to finding the correct coordinates.