Question

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

Answer 1

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

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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;

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

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

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

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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;

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

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

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

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

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On substituting these relations in the formula we get;

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

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

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We know that;

$\sf \hat{i} \times \hat{j} = \hat{j} \times \hat{k} = \sf \hat{k} \times \hat{i} = 0$

‎$\sf \hat{i}^{ \ 2} = \sf \hat{j}^{ \ 2} = \sf \hat{k}^{ \ 2} = 1$

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

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

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

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

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$\sf \dashrightarrow cos\theta = 0$

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We know that cos90° = 0, therefore;

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$\sf \dashrightarrow \theta = 90^\circ$

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Hence proved.