Question

A
vector
is given by
by à = 3i +4j+5k. Find
the magnitude of A ,unit vectors along A
and angles made by
À with coordinate
axis​

Answer 1

Answer:

angle is 45°

Explanation:

What angle does the vector a = 3i - 4j + 5k make with the positive z-axis?

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Here ,

Let b = 0i + 0j + zk ( here the components along x and y is zero because we have to find the angle between a and z-axis) and ¥ be the angle between them

As we know the relation that

a.b= ab cos¥

(3i - 4j + 5k). (0i +0j +zk) = | 3i - 4j + 5k | |0i +0j+ zk| cos ¥

5z = { (3)² +(-4)² +(5)² }½ { z²}½ cos ¥

5 z= √50 z cos ¥

5= 5√2 cos ¥

Cos¥ = 1/√2

Hence ¥ = 45°

So the vector a makes 45° with z-axis

Answer 2

Answer:

1. 45 degree is the answer