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

Practicing the AP Board Solutions Class 11 Maths and Chapter 11 Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11b Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11b

Introduction to 3D Geometry Exercise 11b Solutions

Introduction to 3D Geometry Class 11 Exercise 11b Solutions – Introduction to 3D Geometry 11b Exercise Solutions

I.

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)
Solution:
(i) Distance between the points (2, 3, 5) and (4, 3, 1) = \(\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}\)
= \(\sqrt{4+16}\)
= \(\sqrt{20}\)
= 2√5

(ii) Distance between the points (-3, 7, 2) and (2, 4, -1) = \(\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}\)
= \(\sqrt{25+9+9}\)
= √43

(iii) Distance between the points (-1, 3, -4) and (1, -3, 4) = \(\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}\)
= \(\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}\)
= √104
= 2√26

(iv) Distance between the points (2, -1, 3) and (-2, 1, 3) = \(\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2}\)
= \(\sqrt{16+4}\)
= √20
= 2√5

Question 2.
Find the distance of P(3, -2, 4) from the origin.
Solution:
OP = \(\sqrt{(3-0)^2+(-2-0)^2+(4-0)^2}\)
= \(\sqrt{9+4+16}\)
= √29

Question 3.
Find ‘x’ if the distance between (5, -1, 7) and (x, 5, 1) is 9 units.
Solution:
Let A(5, -1, 7), B(x, 5, 1)
Given AB = 9
⇒ \(\sqrt{(x-5)^2+(5+1)^2+(1-1)^2}\) = 9
⇒ \(\sqrt{x^2-10 x+25+36+36}\) = 9
⇒ x² – 10x + 97 = 81
⇒ x² – 10x + 16 = 0
⇒ (x – 2) (x – 8) = 0
⇒ x = 2 or 8

II.

Question 1.
Show that the points (2, 3, 5), (1, 2, 3), and (7, 0, -1) are collinear.
Solution:
Given A = (-2, 3, 5), B = (1, 2, 3), and C = (7, 0, -1)
\(\overline{\mathrm{AB}}\) = B – A
= (1 – (-2), 2 – 3, 3 – 5)
= (3, -1, -2)
\(\overline{\mathrm{AC}}\) = C – A
= (7 – (-2), 0 – 3, -1 – 5)
= (9, -3, -6)
\(\overline{\mathrm{AC}}\) = K \(\overline{\mathrm{AB}}\)
Take K = 3
3 \(\overline{\mathrm{AB}}\) = 3(3, -1, -2) = (9, -3, -6)
\(\overline{\mathrm{AC}}\) = 3 \(\overline{\mathrm{AB}}\)
Since \(\overline{\mathrm{AC}}\) is a scalar multiple of \(\overline{\mathrm{AB}}\),
The vectors are parallel, so the points A, B, and C are collinear.

Question 2.
Verify the following:
(i) (0, 7, -10), (1, 6, -6), and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (1, 6, 6), and (4, 9, 6) are the vertices of a right-angled triangle.
(iii) (1, 2, 1), (1, -2, 5), (4, -7, 8), and (2, 3, 4) are the vertices of a parallelogram.
Solution:
(i) Let A = (0, 7, -10), B = (1, 6, -6) and C = (4, 9, -6)
Now,

Now AB = BC
∴ The given points are the vertices of an isosceles triangle.

(ii) Let A = (0, 7, 10), B = (-1, 6, 6) and C = (-4, 9, 6)
Now,

Now AB² + BC² = AC²
⇒ ∠B = 90°
∴ The given points are the vertices of a right-angled triangle.

(iii) Let A = (-1, 2, 1), B = (1, -2, 5), C = (4, -7, 8) and D = (2, -3, 4)


AC ≠ BD.
Diagonals are not equal.
∴ ABCD is a parallelogram.

Question 3.
Find the equation of the set of points that are equidistant from the points (1, 2, 3) and (3, 2, -1).
Solution:
Let P(x, y, z) be a point which is equidistant from the points A(1, 2, 3) and B(3, 2, -1).
PA = PB
⇒ PA² = PB²
⇒ (x – 1)² + (y – 2)² +(z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²
⇒ x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 6x + 9 + y² – 4y + 4 + z² + 2z + 1
⇒ -2x – 4y – 6z + 14 = -6x – 4y + 2z + 14
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.

Question 4.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(4, 0, 0) is equal to 10.
Solution:
Let P = (x, y, z), A = (4, 0, 0), B = (-4, 0, 0)
It is given that PA + PB = 10.

Squaring both sides, we obtain
⇒ (x – 4)² + y² + z² = 100 – 20\(\sqrt{(x+4)^2+y^2+z^2}\) + (x + 4)² + y² + z²
⇒ x² – 8x + 16 + y² + z² = 100 – 20\(\sqrt{x^2+8 x+16+y^2+z^2}\) + x² + 8x + 16 + y² + z²
⇒ 20\(\sqrt{x^2+8 x+16+y^2+z^2}\) = 100 + 16x
⇒ 5\(\sqrt{x^2+8 x+16+y^2+z^2}\) = (25 + 4x)
Squaring both sides again, we obtain
⇒ 25(x² + 8x + 16 + y² + z²) = 625 + 16x² + 200x
⇒ 25x² + 200x + 400 + 25y² + 25z² = 625 + 16x² + 200x
⇒ 9x² + 25y² + 25z² – 225 = 0
Thus, the required equation is 9x² + 25y² + 25z² – 225 = 0

Question 5.
If a variable point moves such that 3PA = 2PB, if A = (-2, 2, 3) and B = (13, -3, 13). Prove that P satisfies the equation x² + y² + z² + 28x – 12y + 10z – 247 = 0.
Solution:
Let P = (x, y, z)
Given that A = (-2, 2, 3), B = (13, -3, 13)
3PA = 2PB
⇒ 3\(\sqrt{(x+2)^2+(y-2)^2+(z-3)^2}\) = 2\(\sqrt{(x-13)^2+(y+3)^2+(z-13)^2}\)
⇒ 9(x² + 4x + 4 + y² – 4y + 4 + z² – 6z + 9) = 4(x² – 26x + 169 + y² + 6y + 9 + z² – 26z + 169)
⇒ 9x² + 36x + 36 + 9y² – 36y + 36 + 9z² – 54z + 81 = 4x² + 104x + 676 + 4y² + 24y + 36 + 4z² – 104z + 676
⇒ 5x² + 5y² + 5z² + 140x – 60y + 50z – 1235 = 0
⇒ x² + y² + z² + 28x – 12y + 10z – 247 = 0
∴ P satisfies x² + y² + z² + 28x – 12y + 10z – 247 = 0