Question

If a = (2, -1, 2) and b = (9, 2, 1), find the following. a xb -6i + 16 + 13k x b xa = 6i – 16j – 13k x Need Help? Read It Watch It Talk to a Tutor

2)

3)

Answer 1

So here we have to solve for all the given vector.

__part 1)__

Given vector is;

a = (2, -1,2), b = (9, 2, 1)

We have to find here [
хь
] and [
- X9
] i.e;(Cross product)

So first we find [
хь
].

Let's solve it;

2 j k a xb= 2 -1 -1 2 9 2 1

Expanding;

-1 2. 122. ax6= i +) 21 191 +k 2 -1 19 2

And we get;

a xb=i-1 - 4) + j(18 – 2) + (4 +9)

$a\times b=-5i+16j+13k$

Now we have to find [
- X9
]

2 j k 6 xari 9 2 2 -1 2 1

21 191 bxa=i +) -1 2. 22 +k 19 2 2 - 1

And we get;

b xa=i(4+1) + j(2 – 18) + (-9 – 4)

Therefore we get;

$b\times a=5i-16j-13k$

__Part 2)__

Here we have to find the scalar and vector projection of b onto a.

Given;

a = (4,7,-4), b = (3,-1,1)

Formula for finding the scalar projection of b onto a is;

a.b compa b al

Hence substituting all the values and we get;

compa b 4 - 3 +7-(-1) +(-4) - 1 42 +72 +(-4)2

compa b= 12 – 7 – 4 16 + 49 + 16

1 compa b= 81

compa b=

Now we have to find vector projection of b onto a.

Formula for finding the scalar projection of b onto a is;

a.b proja b = lapz (a)

Substituting all the values and we get;

proja 6 4.3+7.(-1) + (-4): 1 -(4,7,-4) (42 + 72 + (-4)2)

proja b = 12 – 7 - 4 (4,7,-4) (16+49+1602

proja b= (4,7,-4) (812

proja b (9)2 2(4,7, –4)

proja b (4,7,-4) 81

__Part 3)__

Here we have to find the nearest degree of the angles of the triangle with given vertices.

Given vertices are;

P= (1,0), Q = (0,2), R= (5, 4)

And we have to find
ZP, 2,Q, ZR
.

Solving;

1 PQ

(0 - 1), (2-0)

1 PQ

-1,2

$\underset{QR}{\rightarrow}$

(5-01, 4-21

$\underset{QR}{\rightarrow}$

= 5.2

1 PR

- (15-1),(4-01

1 PR

4.4

Now we find the magnitude of
1 PQ
,$\underset{QR}{\rightarrow}$,
1 PR
.

And we get;

1 PQ

= (-1)2 + 22

V5

$\underset{QR}{\rightarrow}$

= V52 + 22 = V29

1 PR

= V42 +42 V32

Now we have to find the angles of the triangle.

For $\angle P$

$\underset{PQ}{\rightarrow},\underset{PR}{\rightarrow};$

$\cos\theta=\dfrac{\underset{PQ}{\rightarrow}\cdot \underset{PR}{\rightarrow}}{|\underset{PQ}{\rightarrow}||\underset{PR}{\rightarrow}|}$

$\theta=\cos^{-1}\dfrac{\underset{PQ}{\rightarrow}\cdot \underset{PR}{\rightarrow}}{|\underset{PQ}{\rightarrow}||\underset{PR}{\rightarrow}|}$

Hence we get;

$\theta=\cos^{-1}\dfrac{-1\cdot 4+2\cdot 4}{\sqrt 5\cdot \sqrt {32}}$

$\theta=\cos^{-1}\dfrac{-4+8}{\sqrt 5\cdot \sqrt {32}}=\cos^{4}\left ( \dfrac{4}{12.64911} \right )=71.57\degree$

Hence we get;

$\angle P=71.57\degree$

Rounding it nearest to 1 decimal place and we get;

Now we finding other angle;

For $\angle Q$

$\underset{PQ}{\rightarrow},\underset{QR}{\rightarrow};$

$\cos\theta=\dfrac{\underset{PQ}{\rightarrow}\cdot \underset{QR}{\rightarrow}}{|\underset{PQ}{\rightarrow}||\underset{QR}{\rightarrow}|}$

$\theta=\cos^{-1}\dfrac{\underset{PQ}{\rightarrow}\cdot \underset{QR}{\rightarrow}}{|\underset{PQ}{\rightarrow}||\underset{QR}{\rightarrow}|}$

Hence we get;

-1.5 +2.2 0 = cos 5 29

$\theta=\cos^{-1}\dfrac{-5+4}{\sqrt 5\cdot \sqrt {29}}=\cos^{-1}\left ( \dfrac{-1}{12.04159} \right )=94.76\degree$

Hence we get;

ZQ = 94.76°

But as it is greater than 90 degree so we find the supplementary angke for this and we get;

$\angle Q=180\degree-94.76\degree=85.24\degree$

Similarly we find the angle $\angle R$

And we get;

$\angle R=23.19\degree$

Rounding it to 1 decimal place and we get;

$\angle R=23.2\degree$

Therefore we get the three angles of the triangle i.e;

$\angle P=71.6\degree,\angle Q=85.2\degree,\angle R=23.2\degree$​​​​​​​