JEE Advance - Mathematics (1998 - No. 2)
For any two vectors $$u$$ and $$v,$$ prove that
(a) $${\left( {u\,.\,v} \right)^2} + {\left| {u \times v} \right|^2} = {\left| u \right|^2}{\left| v \right|^2}$$ and
(b) $$\left( {1 + {{\left| u \right|}^2}} \right)\left( {1 + {{\left| v \right|}^2}} \right) = {\left( {1 - u.v} \right)^2} + {\left| {u + v + \left( {u \times v} \right)} \right|^2}.$$
(a) $${\left( {u\,.\,v} \right)^2} + {\left| {u \times v} \right|^2} = {\left| u \right|^2}{\left| v \right|^2}$$ and
(b) $$\left( {1 + {{\left| u \right|}^2}} \right)\left( {1 + {{\left| v \right|}^2}} \right) = {\left( {1 - u.v} \right)^2} + {\left| {u + v + \left( {u \times v} \right)} \right|^2}.$$
The equation in (a) represents the Pythagorean theorem in a different form.
The equation in (a) relates the dot product and cross product of two vectors to their magnitudes.
The equation in (b) involves the magnitudes, dot product, and cross product of two vectors.
The equation in (b) can be simplified using vector identities.
Both equations are trivial and require no proof.
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