Consider the vector equation of the planes r.(2i + j+k) – 3 and r.(i

1.	Consider the vector equation of the planes r.(2i + j+k) – 3 and r.(i – j – k) = 4 (i) Find the vector equation of any of the plane at the intersection of the above two planes. (ii) Find the vector equation of the planes through the intersection of the above two planes and the point (1, 2, -1)
Answer» (i) 2x + y + z – 3 + λ(x – y – z – 4) = 0 ⇒ (2 + λ)x + (1 – λ)y + (1 – λ)z – 3 – 4λ = 0 Vector equation is . \(\bar{r}\).(2 + λ)i + (1 – λ)y + (1 – λ)k – (3 + 4)λ = 0 (ii) Since passing through (1,2, – 1) we have; ⇒ (2 + λ)1 + (1 – λ)2 + (1 – λ)( -1) – 3 – 4λ = 0 ⇒ 2 + λ + 2 – 2λ – 1 + 1 – 3 – 4λ = 0 ⇒ 0 – 4λ – 0 ⇒ λ = 0 \(\bar{r}\).(2i + j + k) = 3 is required plane. Since the point (1, 2, -1) is a point on the first plane

Consider the vector equation of the planes r.(2i + j+k) – 3 and r.(i – j – k) = 4 (i) Find the vector equation of any of the plane at the intersection of the above two planes. (ii) Find the vector equation of the planes through the intersection of the above two planes and the point (1, 2, -1)

Answer»

(i) 2x + y + z – 3 + λ(x – y – z – 4) = 0

⇒ (2 + λ)x + (1 – λ)y + (1 – λ)z – 3 – 4λ = 0

Vector equation is .

\(\bar{r}\).(2 + λ)i + (1 – λ)y + (1 – λ)k – (3 + 4)λ = 0

(ii) Since passing through (1,2, – 1) we have;

⇒ (2 + λ)1 + (1 – λ)2 + (1 – λ)( -1) – 3 – 4λ = 0

⇒ 2 + λ + 2 – 2λ – 1 + 1 – 3 – 4λ = 0

⇒ 0 – 4λ – 0

⇒ λ = 0

\(\bar{r}\).(2i + j + k) = 3 is required plane. Since the point (1, 2, -1) is a point on the first plane