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Difference of Two Sets

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Question

Find $|\vec{a}| and |\vec{b}|$ if
(i) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 12 and |\vec{a}| = 2 |\vec{b}|$

(ii) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8 and |\vec{a}| = 8 |\vec{b}|$

(iii) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 3 and |\vec{a}| = 2 |\vec{b}|$

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Solution

$(i) Given that |\vec{a}| = 2 |\vec{b}| . . . (1) And (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 12 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 12 \Rightarrow {(2 |\vec{b}|)}^{2} - {|\vec{b}|}^{2} = 12 [From (1)] \Rightarrow 4 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 12 \Rightarrow 3 {|\vec{b}|}^{2} = 12 \Rightarrow {|\vec{b}|}^{2} = 4 \Rightarrow |\vec{b}| = 2 |\vec{a}| = 2 |\vec{b}| = 2 (2) = 4 ∴ |\vec{a}| = 4 and |\vec{b}| = 2$

$(ii) Given that And |\vec{a}| = 8 |\vec{b}| . . . (1) (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 8 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 8 \Rightarrow {(8 |\vec{b}|)}^{2} - {|\vec{b}|}^{2} = 8 [From (1)] \Rightarrow 64 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 8 \Rightarrow 63 {|\vec{b}|}^{2} = 8 \Rightarrow {|\vec{b}|}^{2} = \frac{8}{63} \Rightarrow |\vec{b}| = \sqrt{\frac{8}{63}} |\vec{a}| = 8 |\vec{b}| = 8 \sqrt{\frac{8}{63}} = \frac{8 \sqrt{8}}{\sqrt{63}} ∴ |\vec{a}| = \frac{8 \sqrt{8}}{\sqrt{63}} and |\vec{b}| = \sqrt{\frac{8}{63}}$

$(iii) Given that |\vec{a}| = 2 |\vec{b}| . . . (1) And (\vec{a} + \vec{b}) . (\vec{a} - \vec{b}) = 3 \Rightarrow {|\vec{a}|}^{2} - {|\vec{b}|}^{2} = 3 \Rightarrow {(2 |\vec{b}|)}^{2} - {|\vec{b}|}^{2} = 3 [From (1)] \Rightarrow 4 {|\vec{b}|}^{2} - {|\vec{b}|}^{2} = 3 \Rightarrow 3 {|\vec{b}|}^{2} = 3 \Rightarrow {|\vec{b}|}^{2} = 1 \Rightarrow |\vec{b}| = 1 |\vec{a}| = 2 |\vec{b}| = 2 (1) = 2 ∴ |\vec{a}| = 2 and |\vec{b}| = 1$

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If \vec{a} = \hat{i} - \hat{j} and \vec{b} = - \hat{j} + 2 \hat{k}, find (\vec{a} - 2 \vec{b}) \cdot (\vec{a} + \vec{b}) .

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