Question

If \( \overline{a} \) is a unit vector, then \( |\overline{a} \times \overline{i}|^2 + |\overline{a} \times \overline{j}|^2 + |\overline{a} \times \overline{k}|^2 = \)

• A. $ 4 $
• B. $ 1 $
• C. $ 0 $
• D. $ 2 $

Hint:

When dealing with cross products of unit vectors, use the properties of the cross product and the fact that the sum of squares of the components of a unit vector equals 1.

Answer 1

The Correct Option is D

Step 1: Recall properties of the cross product.
For any vector $ \overline{a} = a_1\overline{i} + a_2\overline{j} + a_3\overline{k} $, the cross products with the standard basis vectors are: $$ \overline{a} \times \overline{i} = a_3\overline{j} - a_2\overline{k}, \quad \overline{a} \times \overline{j} = a_1\overline{k} - a_3\overline{i}, \quad \overline{a} \times \overline{k} = a_2\overline{i} - a_1\overline{j}. $$ The magnitudes squared are: $$ |\overline{a} \times \overline{i}|^2 = (a_3)^2 + (-a_2)^2 = a_3^2 + a_2^2, $$ $$ |\overline{a} \times \overline{j}|^2 = (a_1)^2 + (-a_3)^2 = a_1^2 + a_3^2, $$ $$ |\overline{a} \times \overline{k}|^2 = (a_2)^2 + (-a_1)^2 = a_2^2 + a_1^2. $$ Step 2: Sum the magnitudes squared.
Add these expressions: $$ |\overline{a} \times \overline{i}|^2 + |\overline{a} \times \overline{j}|^2 + |\overline{a} \times \overline{k}|^2 = (a_3^2 + a_2^2) + (a_1^2 + a_3^2) + (a_2^2 + a_1^2). $$ Combine like terms: $$ |\overline{a} \times \overline{i}|^2 + |\overline{a} \times \overline{j}|^2 + |\overline{a} \times \overline{k}|^2 = 2(a_1^2 + a_2^2 + a_3^2). $$ Step 3: Use the fact that $ \overline{a} $ is a unit vector.
Since $ \overline{a} $ is a unit vector, its magnitude is 1: $$ |\overline{a}|^2 = a_1^2 + a_2^2 + a_3^2 = 1. $$ Thus: $$ |\overline{a} \times \overline{i}|^2 + |\overline{a} \times \overline{j}|^2 + |\overline{a} \times \overline{k}|^2 = 2(a_1^2 + a_2^2 + a_3^2) = 2 \cdot 1 = 2. $$ Step 4: Final Answer.
$$ \boxed{2} $$