Question

A vector of magnitude 10 unit perpendicular to a=i+j-k and coplanar with the vector b=2i-j-k and c=i+2j-k is

Answer 1

Answer:

The required vector is

√(50/13) * (- 3i + 4j + k)

Solution:

Let us take a vector

r = xi + yj + zk

Since r = xi + yj + zk and a = i + j - k are perpendicular to each other,

r . a = 0

or, (xi + yj + zk) . (i + j - k) = 0

or, x + y - z = 0 ..... (1)

Since r, b and c vectors are coplanar,

r . (b × c) = 0

| i j k |

or, r . | 2 - 1 - 1 | = 0

| 1 2 - 1 |

or, (xi + yj + zk) . (3i + j + 5k) = 0

or, 3x + y + 5z = 0 ..... (2)

Given magnitude of r = 10 units

This gives √(x² + y² + z²) = 10

or, x² + y² + z² = 100 ..... (3)

Subtracting (1) from (2), we get

2x + 6z = 0

or, x = - 3z

From (1), we get

- 3z + y - z = 0

or, y = 4z

From (3), wet

(- 3z)² + (4z)² + z² = 100

or, 26z² = 100

or, z² = 50/13

or, z = √(50/13)

Then x = - 3√(50/13), y = 4√(50/13)

Therefore, the required vector is

- 3√(50/√13) i + 4√(50/13) j + √(50/13) k

i.e., √(50/13) * (- 3i + 4j + k)