Question

Given:

$\sf \overrightarrow{\sf \ A \ } = 5\hat{i} - \hat{j} - 3\hat{k}$


$\sf \overrightarrow{\sf \ B \ } = \hat{i} + 3\hat{j} - 5\hat{k}$




To prove:

$\sf \overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } \perp\sf \overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }$
And No Spam. ​

Answer 1

Answer:

To prove two vectors to be perpendicular, we must prove that the dot product of those two vectors is equal to zero.

Given:

• A = 5i - j - 3k

• B = i + 3j - 5k

Therefore Vector Addition of A and B is:

⇒ A + B = (5i - j - 3k) + (i + 3j - 5k)

⇒ A + B = (6i + 2j - 8k)

Similarly the value of A - B is:

⇒ A - B = (5i - j - 3k) - (i + 3j - 5k)

⇒ A - B = 5i - j - 3k - i - 3j + 5k

⇒ A - B = (4i - 4j + 2k)

Now to prove (A + B) ⊥ (A - B), we need to prove: (A + B).(A - B) = 0

To perform the dot product of two vectors, we have to multiply corresponding components with each other. That is,

⇒ Component of i × Component of i + Component of j × Component of j

Hence we get:

⇒ (A + B).(A - B) = (6i + 2j - 8k).(4i - 4j + 2k)

⇒ (A + B).(A - B) = (6 × 4) + (2 × -4) + (-8 × 2)

⇒ (A + B).(A - B) = 24 - 8 - 16

⇒ (A + B).(A - B) = 0

Since the dot product of two vectors is zero, the angle between the two vectors is 90°.

Hence (A + B) ⊥ (A - B)

Hence Proved.

Answer 2

Given that ,

${\underline{\boxed{\large{\rm{ \sf \overrightarrow{\sf \ A \ } = 5\hat{i} - \hat{j} - 3\hat{k} }}}}}$

${\underline{\boxed{\large{\rm{\sf \overrightarrow{\sf \ B \ } = \hat{i} + 3\hat{j} - 5\hat{k}}}}}}$

To prove that ,

$\color{brown} \: {\underline{\boxed{\large{\rm{\sf \overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } \perp\sf \overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }}}}}}$

Before proceeding note that :

We know that two non-zero vectors are perpendicular if their scalar product is zero.

____________________________________

Hence , First calculate

$(\sf \overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ })$

$(5\hat{i} - \hat{j} - 3\hat{k} + \hat{i} + 3\hat{j} - 5\hat{k})$

$6\hat{i} + 2\hat{j} - 8\hat{k}</p><p>$

Similarly, now we'll find

$(\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ })$

$5\hat{i} - \hat{j} - 3\hat{k} - (\hat{i} + 3\hat{j} - 5\hat{k})$

$4\hat{i} - 4\hat{j} + 2\hat{k}$

Now , as per the statement we'll prove their scaler product.

$(\sf \overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ })\ (\sf \overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ })$

$24 \: - 8 - \: 16 \:$

$0$

Hence Proved ✔︎✔︎

___________________________________

Hope it helps you please mark it as brainliest if it helps !!