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
Answers
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)