Assertion Let be the vectors such that ^{2}+|\overline{\boldsymbol{b}}|^{2}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}} \) Reason \overline{\boldsymbol{a}}+\overline{\boldsymbol{b}}+\overline{\boldsymbol{a}} \times \overline{\boldsymbol{b}}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}} \)

Both Assertion & Reason are individually true & Reason is correct explanation of Assertion

Maths · Scalar and vector products

Assertion Let be the vectors such that ^{2}+|\overline{\boldsymbol{b}}|^{2}}{\ma

Assertion Let \( \bar{a}, \bar{b}, \bar{r} \) be the vectors such that \( \bar{r}+ \) \( \overline{\boldsymbol{r}} \times \overline{\boldsymbol{a}}=\overline{\boldsymbol{b}} \operatorname{then}|\overline{\boldsymbol{r}}|^{2}=\frac{(\overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{b}})^{2}+|\overline{\boldsymbol{b}}|^{2}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}} \) Reason \( \overline{\boldsymbol{r}}=\frac{(\overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{b}}) \overline{\boldsymbol{a}}+\overline{\boldsymbol{b}}+\overline{\boldsymbol{a}} \times \overline{\boldsymbol{b}}}{\mathbf{1}+|\overline{\boldsymbol{a}}|^{2}} \)

A. Both Assertion \& Reason are individually true \& Reason is correct explanation of Assertion
B. Both Assertion \& Reason are individually true but Reason is not the correct (proper) explanation of Assertion
C. Assertion is true but Reason is false
D. Assertion is false but Reason is true

Step-by-step solution

From the given equation r + r×a = b, taking dot product with a gives a·r = a·b. Assuming r = αa + βb + γ(a×b) and substituting yields α = (a·b)/(1+|a|^2), β = γ = 1/(1+|a|^2), leading to r = ((a·b)a + b + a×b)/(1+|a|^2). Then |r|^2 = ((a·b)^2+|b|^2)/(1+|a|^2). Thus both statements are true and the reason provides the correct derivation.

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Step-by-step solution

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