Question

Find (a) scalar component and (b) vector component of vector A=3i+4j+5k on vector B=i+j+k.

Answer 1

Answer:

Scalar component of vector A=

(3i^ + 4j^+ 5k^).(i^+j^+k^)

A=3+4+5 [Since i with j with k value becomes 0]

So,Vector A=12

Vector component vector A=

(Using determinant,cross product or CRAEMER's rule)

Vector A=4-5+(5-3)+(3-4)

1+2-1

Vector A=2

HOPE IT HELPS... ☺️☺️

Answer 2

The correct answers are a) 4√3 b) 4i + 4j + 4k.

Given:

Two vectors, $\vec A$  = 3i + 4j + 5k and $\vec B$ = i + j + k.

To Find:

The (a) scalar component and (b) vector component of the vector

$\vec A$  = 3i+4j+5k on the vector $\vec B$ = i+j+k.

Solution:

In this question, we need to find the scalar and vector components of a vector $\vec A$ , when it is on another vector $\vec B$.

To solve this question we will use the following formulae:

i) Scalar component of the vector $\vec A$ on vector $\vec B$ is  = $\frac{\vec A. \vec B}{|\vec B|}$

ii) Vector  component of vector $\vec A$ on vector $\vec B$ is  = $\frac{\vec A. \vec B}{|\vec B|}$

We have been given two vectors, $\vec A$  = 3i + 4j + 5k and $\vec B$ = i + j + k.

Now,

1) The dot product of the given two vectors is

$\vec A$ . $\vec B$ = (3i + 4j + 5k).(i + j + k) = 3 + 4 + 5 = 12

2) The magnitude of the vector $\vec B$, | $\vec B$ | = $\sqrt{1^{2} +1^{2} +1^{2} }$ = √3

3) The unit vector of  $\vec B$, $\hat{B}$ = $\frac{ \vec B}{|\vec B|}$ = $\frac{(\hat{i}+ \hat{j} + \hat{k)}}{\sqrt{3} }$

 Now,

a) Scalar component of the vector $\vec A$ on vector $\vec B$ is  = $\frac{\vec A. \vec B}{|\vec B|}$ = 12/√3 = 4√3.

b) Vector  component of vector $\vec A$ on vector $\vec B$ is  = $\frac{\vec A. \vec B}{|\vec B|}$ $(\hat{B})$

= $\frac{12}{\sqrt{3} }$ x $\frac{(\hat{i}+ \hat{j} + \hat{k)}}{\sqrt{3} }$ = 4(i + j + k) = 4i + 4j + 4k.

Hence, the correct answers are a) 4√3 b) 4i + 4j + 4k.

#SPJ3