Question

35. If A = 2i – 3j + 4k, the magnitudes of its components in yz-plane and zx-plane are respectively

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Answer 1

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Answer 2

Concept:

We need to first recall the concept of vector and magnitude of its component to answer this question.

• Vector is the quantity which has both magnitude and direction.

• The magnitude of a vector is the length of the vector v denoted by ||v||.

Given:

A vector A = 2i-3j+4k

To find:

The magnitudes of components of vector A in yz- plane and zx-plane.

Solution:

The vector given is:

A = 2i-3j+4k

On comparing with general form A= xi+yj+zk

we get x = 2 , y = -3 and z = 4

Along yz-plane

Magnitude of component is given by : $\sqrt{y^{2}+z^{2}}$

                                                            = $\sqrt{(-3)^{2}+(4)^{2}}$

                                                            = $\sqrt{9+16}$

                                                            = $\sqrt{25}$

                                                           = 5

Along zx-plane

Magnitude of component is given by : $\sqrt{z^{2}+x^{2}}$

                                                            = $\sqrt{(4)^{2}+(2)^{2}}$

                                                            = $\sqrt{16+4}$

                                                            = $\sqrt{20}$

                                                            = 2$\sqrt{5}$

Hence the magnitude along yz plane is 5 and along zx plane is 2$\sqrt{5}$.