For any vector {{ a}} {≤ft | {{{ a}} × {{ i}}} | ² + {≤ft | {{{ a}} × {{ j}}} | ² + {≤ft | {{{ a}} × {{ k}}} | ² is equal to

Correct Answer is : 2∣a⃗∣22{≤ft | {{{ a}}} | ²2∣a∣2

Correct Answer is : 2∣a⃗∣22{≤ft | {{{ a}}} | ²2∣a∣2

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 48 of 120

Marks: +1, -0

For any vector

{\rm{\vec a}}

{\left | {{\rm{\vec a}} \times {\rm{\hat i}}} \right | ^2} + {\left | {{\rm{\vec a}} \times {\rm{\hat j}}} \right | ^2} + {\left | {{\rm{\vec a}} \times {\rm{\hat k}}} \right | ^2}

is equal to

Solution:

Concept:

The sum of squares of magnitudes of cross products of a vector with unit vectors

\hat i

\hat j

\hat k

equals twice the square of its magnitude.

Explanation:

Let

\vec a = x\hat i + y\hat j + z\hat k

Compute each cross product using determinant form:

\vec a \times \hat i = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & z \\ 1 & 0 & 0 \end{vmatrix} = z\hat j - y\hat k

Its magnitude squared is

|\vec a \times \hat i|^2 = z^2 + y^2

Similarly,

\vec a \times \hat j = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & z \\ 0 & 1 & 0 \end{vmatrix} = -z\hat i + x\hat k

, so

|\vec a \times \hat j|^2 = z^2 + x^2

And

\vec a \times \hat k = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & z \\ 0 & 0 & 1 \end{vmatrix} = y\hat i - x\hat j

, so

|\vec a \times \hat k|^2 = y^2 + x^2

Add the three expressions:

(z^2+y^2)+(z^2+x^2)+(y^2+x^2) = 2(x^2+y^2+z^2) = 2|\vec a|^2

Answer:

Option B:

2|\vec a|^2

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