Find α given that 5i+4j and αj−3k have the same length.

The value of α = 4√2

The length of a vector r = xi + yj + zk with x and y component and z components , x and y  and z is |r| = √(x² + y² + z²)

Let a = 5i + 4j and b = αj − 3k

So, length of a is |a| = √(5² + 4² + 0²)

= √(25 + 16)

= √41

length of b is |b| = √(0² + α² + (-3)²)

= √(α² + 9)

Since |a| = |b|

√41 = √(α² + 9)

Squaring both sides, we have

41 = α² + 9

α² = 41 - 9

α² = 32

Taking square root of both sides, we have

α = √32

α = √(16 × 2)

α = 4√2

So, the value of α = 4√2

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