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Polynomials

For any non-z...

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For any non-zero vector $\vec{r}, {|\vec{r} \times \hat{i}|}^{2} + {|\vec{r} \times \hat{j}|}^{2} + {|\vec{r} \times \hat{k}|}^{2} = λ {|\vec{r}|}^{2}, then λ =___________________.$

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Solution

$Let \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} Then \vec{r} \times \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{i} = x (\hat{i} \times \hat{i}) + y (\hat{j} \times \hat{i}) + z (\hat{k} \times \hat{i}) i . e \vec{r} \times \hat{i} = 0 - y \hat{k} + z \hat{j} . . . (1) (∵ \hat{j} \times \hat{i} = - \hat{k} and \hat{k} \times \hat{i} = \hat{j}) Similarly, \vec{r} \times \hat{k} and \vec{r} \times \hat{j} are calculated i . e . \vec{r} \times \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{j} = x (\hat{i} \times \hat{j}) + y (\hat{j} \times \hat{j}) + z (\hat{k} \times \hat{j}) = x \hat{k} + y (0) - z \hat{i} i . e . \vec{r} \times \hat{j} = x \hat{k} - z \hat{i} . . . (2) and \vec{r} \times \hat{k} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{k} = x (\hat{i} \times \hat{k}) + y (\hat{j} \times \hat{k}) + z (\hat{k} \times \hat{k}) = - x \hat{j} + y \hat{i} + z (0) i . e . \vec{r} \times \hat{k} = - x \hat{j} + y \hat{i} . . . (3) ∴ From (1), (2) and (3) {|\vec{r} \times \hat{i}|}^{2} + {|\vec{r} \times \hat{j}|}^{2} + {|\vec{r} \times \hat{k}|}^{2} = {|- y \hat{k} + z \hat{j}|}^{2} + {|x \hat{k} - z \hat{i}|}^{2} + {|- x \hat{j} + y \hat{i}|}^{2} = y^{2} + z^{2} + x^{2} + z^{2} + x^{2} + y^{2} = 2 (x^{2} + y^{2} + z^{2}) = 2 {|r|}^{2} = λ {|r|}^{2} (given) ∴ λ = 2$

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