OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a)

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(a)

Question 1.
Find the distance from the origin to each of the points :
(i) (2, 2, 3)
(ii) (4, -1, 2)
(iii) (0, 4, -4)
(iv) (-4, -3, -2)
Solution:
(i) Required distance of origin O(0, 0, 0) from P(2, 2, 3) = \(\sqrt{(2-0)^2+(2-0)^2+(3-0)^2}\) = \(\sqrt{4+4+9}\) = \(\sqrt{17}\)
(ii) Required distance =\(\sqrt{(4-0)^2+(-1-0)^2+(2-0)^2}\) = \(\sqrt{16+1+4}\) = \(\sqrt{21}\)
(iii) Required distance = \(\sqrt{(0-0)^2+(4-0)^2+(-4-0)^2}\) = \(\sqrt{16+16}\) = \(4 \sqrt{2}\)
(iv) Required distance = \(\sqrt{(-4-0)^2+(-3-0)^2+(-2-0)^2}\) = \(\sqrt{16+9+4}\) = \(\sqrt{29}\)

Question 2.
Find the distance between each of the following pairs of points :
(i) (2, 5, 3) and (-3, 2, 1);
(ii)(0, 3, 0) and (6, 0, 2);
(iii) (-4, -2, 0) and (3, 3, 5).
Solution:
(i) Required distance = \(\sqrt{(-3-2)^2+(2-5)^2+(1-3)^2}\) = \(\sqrt{25+9+4}\) = \(\sqrt{38}\)
(ii) Required distance = \(\sqrt{(6-0)^2+(0-3)^2+(2-0)^2}\) = \(\sqrt{36+9+4}\) = \(\sqrt{49}\) = 7
(iii) Required distance = \(\sqrt{(3+4)^2+(3+2)^2+(5-0)^2}\) = \(\sqrt{49+25+25}\) = \(\sqrt{99}\) = \(3 \sqrt{11}\)

Question 3.
Show that the triangle with vertices (6, 10, 10),(1, 0, -5),(6, -10, 0) is a right-angled triangle, and find its axes.
Solution:
Let the given vertices of triangle are A(6, 10, 10); B(1, 0, -5) and C (6, -10, 0)
∴ AB = \(\sqrt{(1-6)^2+(0-10)^2+(-5-10)^2}\) = \(\sqrt{25+100+225}\) = \(\sqrt{350}\)
BC = \(\sqrt{(6-1)^2+(-10-0)^2+(0+5)^2}\) = \(\sqrt{25+100+25}\) = \(\sqrt{150}\)
CA = \(\sqrt{(6-6)^2+(-10-10)^2+(0-10)^2}\) = \(\sqrt{400+100}\) = \(\sqrt{500}\)
∴ AB² + BC² =350 + 150 = 500 = CA²
Thus △ABC be right angled △ at B.
Therefore A, B and C are the vertices of right angled triangle.
∴ area of △ABC = \(\frac { 1 }{ 2 }\) (AB) × (BC) = \(\frac { 1 }{ 2 }\) × \(\sqrt{350} \sqrt{150}\) = \(\frac { 1 }{ 2 }\) \(\sqrt{35 \times 15 \times 100}\)
= \(\frac { 1 }{ 2 }\) × 50√21 = 25√21 sq. units

Question 4.
Show that the triangle with vertices A(3, 5, -4), B(-1, 1, 2), C(-5, -5, -2) is isosceles.
Solution:
Given vertices of triangle are A (3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2).
AB = \(\sqrt{(-1-3)^2+(1-3)^2+(2+4)^2}\) = \(\sqrt{16+16+36}\) = \(\sqrt{68}\) = \(2 \sqrt{17}\)
BC = \(\sqrt{(-5+1)^2+(-5-1)^2+(-2-2)^2}\) = \(\sqrt{16+36+16}\) = \(\sqrt{68}\)
CA = \(\sqrt{(3+5)^2+(5+5)^2+(-4+2)^2}\) = \(\sqrt{64+100+4}\) = \(\sqrt{168}\)
Thus AB = BC
∴ △ ABC be an isosceles triangle.

Question 5.
Show that (4, 2, 4),(10, 2, -2) and (2, 0, -4) are the vertices of an equilateral triangle.
Solution:
Let the vertices of triangle are A(4, 2, 4), B(10, 2, -2) and C(2, 0, -4).
AB = \(\sqrt{(10-4)^2+(2-2)^2+(-2-4)^2}\) = \(\sqrt{36+36}\) = \(6 \sqrt{2}\) = \(\sqrt{72}\)
BC = \(\sqrt{(2-10)^2+(0-2)^2+(-4+2)^2}\) = \(\sqrt{64+4+4}\) = \(\sqrt{72}\)
and CA = \(\sqrt{(4-2)^2+(2-0)^2+(4+4)^2}\) = \(\sqrt{4+4+64}\) = \(\sqrt{72}\)
Thus AB = BC = CA
∴ △ ABC is an equilateral triangle.

Question 6.
Show that the points (1, -1, 3),(2, -4, 5) and (5, -13, 11) are collinear.
Solution:
Given points are A(1, -1, 3); B(2, -4, 5) and C(5, -13, 11)
Here AB = \(\sqrt{(2-1)^2+(-4+1)^2+(5-3)^2}\) = \(\sqrt{1+9+4}\) = \(\sqrt{14}\)
BC = \(\sqrt{(5-2)^2+(-13+4)^2+(11-5)^2}\) = \(\sqrt{9+81+36}\) = \(\sqrt{126}\) = \(3 \sqrt{14}\)
CA = \(\sqrt{(1-5)^2+(-1+13)^2+(3-11)^2}\) = \(\sqrt{16+144+64}\) = \(\sqrt{224}\) = \(4 \sqrt{14}\)
Clearly AB + BC = CA
Thus the points A, B and C are collinear.

Question 7.
Derive the equation of the locus of a point equidistant from the points (1, -2, 3) and (-3, 4, 2).
Solution:
Let P(x, y, z) be any point on locus and A(1, -2, 3), B(-3, 4, 2) are given points.
According to given condition, we have |PA| = | PB |
\(\sqrt{(x-1)^2+(y+2)^2+(z-3)^2}\) = \(\sqrt{(x+3)^2+(y-4)^2+(z-2)^2}\)
On squaring both sides; we have
(x – 1)² + (y + 2)² + (z – 3)² =(x + 3)² + (y – 4)² + (z – 2)²
⇒ -2x +4y – 6z + 14 = 6x – 8y – 4z + 29
⇒ 8x – 12y + 2z + 15 = 0
which is the required locus of a point.

Question 8.
Derive the equation of the locus of a point twice as far from (-2, 3, 4) as from (3, -1, -2).
Solution:
Let P(x, y, z) be any point on the locus
such that PA = 2 PB
where A be the point (-2, 3, 4) and B be the point (+3, -1, -2).
\(\sqrt{(x+2)^2+(y-3)^2+(z-4)^2}\) = \(2 \sqrt{(x-3)^2+(y+1)^2+(z+2)^2}\)
On squaring both sides; we have
(x + 2)² + (y – 3)² + (z – 4)² =4[(x – 3)² + (y + 1)² + (z + 2)²]
⇒ 3x² + 3y² + 3z² – 28x + 14y + 24z + 27 = 0
which is the required locus.

Question 9.
Find the equation of the locus of a point whose distance from the y-axis is equal to its distance from (2, 1, -1).
Solution:
Let P(x, y, z) be any point on locus and A(0, y, 0) be any point on y-axis and B(2, 1, -1) be the given point.
According to given condition, PA = PB
\(\sqrt{(x-0)^2+(y-y)^2+(z-0)^2}\) = \(\sqrt{(x-2)^2+(y-1)^2+(z+1)^2}\)
On squaring both sides; we have
x² + z² = x² + y² + z² – 4x – 2y + 2z + 6
⇒ y² – 4x – 2y + 2z + 6 = 0 be the required eqn. of locus.

Question 10.
Find the equation of the locus of a point whose distance from the xy-plane is equal to distance from the point (-1, 2, -3).
Solution:
Let P(x, y, z) be any point on locus and A(x, y, 0) be any point in xy plane and B(-1, 2, -3) b the given point.
Then according to given condition; we have PA = PB
\(\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}\) = \(\sqrt{(x+1)^2+(y-2)^2+(z+3)^2}\)
On squaring both sides; we have
0 + 0 + z² = x² + y² + z² + 2x – 4y + 6z + 14
⇒ x² + y² + 2x – 4y + 6z + 14 = 0
which is the required locus.

Question 11.
A point moves so that the difference of the squares of its distances from the x-axis and the y-axis is constant. Find the equatiol of its locus.
Solution:
Let P(x, y, z) be any point on locus and let Q(x, 0, 0) and R(0, y, 0) be any two points on x-axis and y-axis.
Then according to given condition; we have
PQ² – PR² = constant = k
⇒ \(\left[\sqrt{(x-x)^2+(y-0)^2+(z-0)^2}\right]^2\) – \(\left[\sqrt{(x-0)^2+(y-y)^2+(z-0)^2}\right]^2\) = k
⇒ y² + z² – x² – z² = k
⇒ y² – x² = k
which is the required locus.

Question 12.
Find the equation of the locus of a point whose distance from the z-axis is equal to its distance from the xy-plane.
Solution:
Let P(x, y, z) be any point on locus and let Q(0, 0, z) be any point on z-axis and R(x, y, 0) be any point in xy-plane.
Then PQ = PR
⇒ \(\sqrt{(x-0)^2+(y-0)^2+(z-z)^2}\) = \(\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}\)
On squaring both sides; we have
x² + y² = z²
⇒ x² + y² – z² = 0
which is the required locus.