Question

Deduce the condition for the vectors 2hati + 3hatj - 4hatk and 3hati - alpha hatj + bhatk to be parallel.

Answer 1

The condition for the vectors 2hati + 3hatj - 4hatk and 3hati - alpha hatj + bhatk to be parallel is α = -9/2 and b = -6.

For two vectors to be parallel, the unit vector along both the vectors should be same.

Therefore unit vector along first vector = V/|V| = 2/√29i + 3/√29j - 4/√29k

Unit vector along second vector = (3i - αj + bk)/√9 + α² + b²

Equating i , j and k components,

• 3/√9 + α² + b² = 2/√29

• 9 x 29 = 4 x( 9 + α² + b² )

• 261 = 36 + 4α² + 4b²

• α² + b² = 225/4 - - (1) x 9 == >
• 9α² + 9b² = 2025/4 --(2)

• 3/√29 = -α/√9 + α² + b²
• 81 + 9α² + 9b² = 29α²
• 20α² - 9b² = 81 (3)

( 2) + (3)

• 29α² = 2349/4
• α² = 81/4
• α = ±9/2
• We take α = -9/2 , since j component of first vector is positive.

From (1)

• b² = 225/4 - α² = 225 - 81 / 4 = 144/4
• b = ±12/2
• We take b  = -6 ,  since k component of first vector is negative