JEE MAIN - Mathematics (2005 - No. 54)
For any vector $${\overrightarrow a }$$ , the value of $${\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2}$$ is equal to :
$$3{\overrightarrow a ^2}$$
$${\overrightarrow a ^2}$$
$$2{\overrightarrow a ^2}$$
$$4{\overrightarrow a ^2}$$
Explanation
Let $$\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $$
$$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $$
$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$$
Similarly, $${\left( {\overrightarrow a \times \overrightarrow j } \right)^2} = {x^2} + {z^2}\,\,$$
and $${\left( {\overrightarrow a \times \overrightarrow k } \right)^2} = {x^2} + {y^2}$$
$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow j } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow k } \right)^2}$$
$$ = 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2\overrightarrow a 2$$
$$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $$
$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$$
Similarly, $${\left( {\overrightarrow a \times \overrightarrow j } \right)^2} = {x^2} + {z^2}\,\,$$
and $${\left( {\overrightarrow a \times \overrightarrow k } \right)^2} = {x^2} + {y^2}$$
$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow j } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow k } \right)^2}$$
$$ = 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2\overrightarrow a 2$$
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