If the distance between the points P(a, 2, 1) and Q(1, -1, 1) is 5 uni

1.	If the distance between the points P(a, 2, 1) and Q(1, -1, 1) is 5 units, find the value of a.
Answer» Given: distance between the points P(a, 2, 1) and Q(1, -1, 1) is 5 units To find: the value of a Formula used: The distance between any two points (a, b, c) and (m, n, o) is given by, \(\sqrt{(a-m)^2+(b-n)^2+(c-o)^2}\) PQ = 5 units The distance between points P(a, 2, 1) and Q(1, -1, 1) is PQ ⇒\(\sqrt{(a-1)^2+(2-(-1))^2+(1-1)^2}\) ⇒\(\sqrt{(a-1)^2+3^2+0^2}\) = 5 ⇒ \(\sqrt{a^2+1-2a+9+0}\) = 5 ⇒ \(\sqrt{a^2-2a+10}\) = 5 Squaring both sides: ⇒a² – 2a + 10 = 25 ⇒ a²– 2a + 10 – 25 = 0 ⇒ a² – 2a – 15 = 0 ⇒ a² – 5a + 3a – 15 = 0 ⇒ a(a – 5) + 3(a – 5) = 0 ⇒ (a – 5) (a + 3) = 0 ⇒ a = 5 or -3 Hence, the value of a is 5 or -3

If the distance between the points P(a, 2, 1) and Q(1, -1, 1) is 5 units, find the value of a.

Answer»

Given: distance between the points P(a, 2, 1) and Q(1, -1, 1) is 5 units

To find: the value of a

Formula used: The distance between any two points (a, b, c) and (m, n, o) is given by,

\(\sqrt{(a-m)^2+(b-n)^2+(c-o)^2}\)

PQ = 5 units

The distance between points P(a, 2, 1) and Q(1, -1, 1) is PQ

⇒\(\sqrt{(a-1)^2+(2-(-1))^2+(1-1)^2}\)

⇒\(\sqrt{(a-1)^2+3^2+0^2}\) = 5

⇒ \(\sqrt{a^2+1-2a+9+0}\) = 5

⇒ \(\sqrt{a^2-2a+10}\) = 5

Squaring both sides:

⇒a² – 2a + 10 = 25

⇒ a²– 2a + 10 – 25 = 0

⇒ a² – 2a – 15 = 0

⇒ a² – 5a + 3a – 15 = 0

⇒ a(a – 5) + 3(a – 5) = 0

⇒ (a – 5) (a + 3) = 0

⇒ a = 5 or -3

Hence, the value of a is 5 or -3