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if a = 5i - j -3k and b= i+3j-5k, then show that the vectors a+b and a-b are perpendicular

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Answer 1

Given:

• a = 5î - ĵ - 3k
• b = î + 3ĵ - 5k

To Show:

• (a + b) and (a - b) are perpendicular

Solution:

Firstly let us find the values of (a + b) and (a - b) individually!

For (a + b)

(a + b) = (5î - ĵ - 3k) + (î + 3ĵ - 5k)

(a + b) = (5 + 1)î + (-1 + 3)ĵ + (-3 -5)k

(a + b) = 6î + 2ĵ - 8k

For (a - b)

(a - b) = (5î - ĵ - 3k) - (î + 3ĵ - 5k)

(a - b) = (5 - 1)î + (-1 -3)ĵ + (-3 + 5)k

(a - b) = 4î - 4ĵ + 2k

Now,

Now,We know:

• If the scalar product of (a + b) and (a - b) comes to be zero, it means both are Perpendicular to each other.

Therefore:

Scalar product = (a + b) . (a - b)

(a + b) . (a - b) = (6î + 2ĵ - 8k) . (4î - 4ĵ + 2k)

(a + b) . (a - b) = (6 × 4) + {2 × (-4)} + {(-8) × 2}

(a + b) . (a - b) = 24 - 8 - 16

(a + b) . (a - b) = 24 - 24

(a + b) . (a - b) = 0

Here, (a + b) . (a - b) = 0 and we know if scalar product of two vectors is 0 then the two vectors are Perpendicular to each other.

Hence, (a + b) and (a - b) are perpendicular [Proved]

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Answer 2

Answer:

Given:

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

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

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To prove:

$\sf \overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } \perp\sf \overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }$

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Solution:

For any two vectors to be perpendicular to each other, the angle formed between them should 90°.

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The angle between any two vectors is given by the dot product of the two vectors divided by the product of the magnitude of both the vectors.

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$\begin{gathered}\sf \dashrightarrow cos\theta = \dfrac{\left(\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ }\right)\bullet \left(\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }\right)}{\left|\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ }\right| \left|\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }\right|} \\ \\\end{gathered}$

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$Calculating _{\overrightarrow{\ \sf A \ } + \overrightarrow{\ \sf B \ }}$

:

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

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

$\sf \dashrightarrow \underline{\underline{\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } = 6\hat{i} + 2\hat{j} - 8\hat{k}}} \ \dots \ Relation (1)$

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$Calculating _{\overrightarrow{\ \sf A \ } - \overrightarrow{\ \sf B \ }}$

:

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

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

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

$\sf \dashrightarrow \underline{\underline{\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ } = 4\hat{i} - 4\hat{j} + 2\hat{k}}} \ \dots \ Relation (2)$

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$Calculating _{| \overrightarrow{\ \sf A \ } + \overrightarrow{\ \sf B \ }|}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } | = \sqrt{x^2 + y^2 + z^2}$

On substituting the value obtained in Relation (1) we get;

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } | = \sqrt{(6)^2 + (2)^2 + (-8)^2}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } | = \sqrt{36 + 4 + 64}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ } | = \sqrt{104} \ \dots \ Relation (3) … Relation(3)$

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$Calculating _{| \overrightarrow{\ \sf A \ } - \overrightarrow{\ \sf B \ }|}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ } | = \sqrt{x^2 + y^2 + z^2}$

On substituting the value obtained in Relation (2) we get;

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ } | = \sqrt{(4)^2 + (-4)^2 + (2)^2}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ } | = \sqrt{16 + 16 + 4}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ } | = \sqrt{36}$

$\sf \dashrightarrow |\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ } | = 6 \ Relation(4)$

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On substituting these relations in the formula we get;

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$\begin{gathered}\sf \dashrightarrow cos\theta = \dfrac{\left(\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ }\right)\bullet \left(\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }\right)}{\left|\overrightarrow{\sf \ A \ } + \sf \overrightarrow{\sf \ B \ }\right| \left|\overrightarrow{\sf \ A \ } - \sf \overrightarrow{\sf \ B \ }\right|} \\ \\\end{gathered}$

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$\begin{gathered}\sf \dashrightarrow cos\theta = \dfrac{\left(6\hat{i}\right)\left(4\hat{i}\right) + \left(2\hat{j}\right)\left(- 4\hat{j}\right) - \left(8\hat{k}\right)\left(2\hat{k}\right)}{\sqrt{104} \times 6} \\ \\\end{gathered}$

$\begin{gathered}\sf \dashrightarrow cos\theta = \dfrac{24 - 8 - 16}{\sqrt{104} \times 6} \\ \\\end{gathered}$

$\begin{gathered}\sf \dashrightarrow cos\theta = \dfrac{16 - 16}{\sqrt{104} \times 6} \\ \\\end{gathered}$

$\begin{gathered}\sf \dashrightarrow cos\theta = \dfrac{0}{\sqrt{104} \times 6} \\ \\\end{gathered}$

$\sf \dashrightarrow cos\theta = 0⇢cosθ=0$

$We \: know \: that \: cos90° = 0, \: therefore;$

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$\sf \dashrightarrow \theta = 90^\circ⇢θ=90 {}^{0}$

vectors a+b and a-b are perpendicular

Hence proved.