Key Concepts and Formulas

• Vector Cross Product: The cross product of two vectors $\overrightarrow u$ and $\overrightarrow v$, denoted by $\overrightarrow u \times \overrightarrow v$, results in a vector perpendicular to both $\overrightarrow u$ and $\overrightarrow v$. Its magnitude is given by $|\overrightarrow u \times \overrightarrow v| = |\overrightarrow u| |\overrightarrow v| \sin \theta$, where $\theta$ is the angle between the vectors.
• Unit Vectors: $\widehat i$, $\widehat j$, and $\widehat k$ are mutually orthogonal unit vectors along the x, y, and z axes, respectively. We have $\widehat i \times \widehat j = \widehat k$, $\widehat j \times \widehat k = \widehat i$, $\widehat k \times \widehat i = \widehat j$, and $\widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = \overrightarrow 0$. Also, $\widehat i \times \widehat k = -\widehat j$, etc.
• Magnitude of a Vector: The square of the magnitude of a vector $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$ is $|\overrightarrow a|^2 = \overrightarrow a \cdot \overrightarrow a = x^2 + y^2 + z^2$.

Step-by-Step Solution

Let the given vector be $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$. We need to find the value of ${\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2}$.

Step 1: Calculate the cross product $\overrightarrow a \times \widehat i$. We use the distributive property of the cross product and the properties of unit vectors. $\overrightarrow a \times \widehat i = (x\widehat i + y\widehat j + z\widehat k) \times \widehat i$ $= (x\widehat i \times \widehat i) + (y\widehat j \times \widehat i) + (z\widehat k \times \widehat i)$ Since $\widehat i \times \widehat i = \overrightarrow 0$, $\widehat j \times \widehat i = -\widehat k$, and $\widehat k \times \widehat i = \widehat j$, we get: $\overrightarrow a \times \widehat i = x(\overrightarrow 0) + y(-\widehat k) + z(\widehat j)$ $\overrightarrow a \times \widehat i = -y\widehat k + z\widehat j$

Step 2: Calculate the square of the magnitude of $\overrightarrow a \times \widehat i$. The square of the magnitude of a vector is the dot product of the vector with itself. $\left( {\overrightarrow a \times \widehat i} \right)^2 = (-y\widehat k + z\widehat j) \cdot (-y\widehat k + z\widehat j)$ $= (-y\widehat k) \cdot (-y\widehat k) + (-y\widehat k) \cdot (z\widehat j) + (z\widehat j) \cdot (-y\widehat k) + (z\widehat j) \cdot (z\widehat j)$ Since $\widehat j \cdot \widehat k = 0$, $\widehat k \cdot \widehat j = 0$, and $\widehat j \cdot \widehat j = 1$, $\widehat k \cdot \widehat k = 1$: $\left( {\overrightarrow a \times \widehat i} \right)^2 = (-y)^2 (\widehat k \cdot \widehat k) + 0 + 0 + (z)^2 (\widehat j \cdot \widehat j)$ $\left( {\overrightarrow a \times \widehat i} \right)^2 = y^2 (1) + z^2 (1)$ $\left( {\overrightarrow a \times \widehat i} \right)^2 = y^2 + z^2$

Step 3: Calculate the cross product $\overrightarrow a \times \widehat j$. Using the distributive property and the properties of unit vectors: $\overrightarrow a \times \widehat j = (x\widehat i + y\widehat j + z\widehat k) \times \widehat j$ $= (x\widehat i \times \widehat j) + (y\widehat j \times \widehat j) + (z\widehat k \times \widehat j)$ Since $\widehat i \times \widehat j = \widehat k$, $\widehat j \times \widehat j = \overrightarrow 0$, and $\widehat k \times \widehat j = -\widehat i$: $\overrightarrow a \times \widehat j = x(\widehat k) + y(\overrightarrow 0) + z(-\widehat i)$ $\overrightarrow a \times \widehat j = x\widehat k - z\widehat i$

Step 4: Calculate the square of the magnitude of $\overrightarrow a \times \widehat j$. $\left( {\overrightarrow a \times \widehat j} \right)^2 = (x\widehat k - z\widehat i) \cdot (x\widehat k - z\widehat i)$ $= (x\widehat k) \cdot (x\widehat k) + (x\widehat k) \cdot (-z\widehat i) + (-z\widehat i) \cdot (x\widehat k) + (-z\widehat i) \cdot (-z\widehat i)$ Since $\widehat i \cdot \widehat k = 0$, $\widehat k \cdot \widehat i = 0$, and $\widehat i \cdot \widehat i = 1$, $\widehat k \cdot \widehat k = 1$: $\left( {\overrightarrow a \times \widehat j} \right)^2 = x^2 (\widehat k \cdot \widehat k) + 0 + 0 + (-z)^2 (\widehat i \cdot \widehat i)$ $\left( {\overrightarrow a \times \widehat j} \right)^2 = x^2 (1) + z^2 (1)$ $\left( {\overrightarrow a \times \widehat j} \right)^2 = x^2 + z^2$

Step 5: Calculate the cross product $\overrightarrow a \times \widehat k$. Using the distributive property and the properties of unit vectors: $\overrightarrow a \times \widehat k = (x\widehat i + y\widehat j + z\widehat k) \times \widehat k$ $= (x\widehat i \times \widehat k) + (y\widehat j \times \widehat k) + (z\widehat k \times \widehat k)$ Since $\widehat i \times \widehat k = -\widehat j$, $\widehat j \times \widehat k = \widehat i$, and $\widehat k \times \widehat k = \overrightarrow 0$: $\overrightarrow a \times \widehat k = x(-\widehat j) + y(\widehat i) + z(\overrightarrow 0)$ $\overrightarrow a \times \widehat k = -x\widehat j + y\widehat i$

Step 6: Calculate the square of the magnitude of $\overrightarrow a \times \widehat k$. $\left( {\overrightarrow a \times \widehat k} \right)^2 = (-x\widehat j + y\widehat i) \cdot (-x\widehat j + y\widehat i)$ $= (-x\widehat j) \cdot (-x\widehat j) + (-x\widehat j) \cdot (y\widehat i) + (y\widehat i) \cdot (-x\widehat j) + (y\widehat i) \cdot (y\widehat i)$ Since $\widehat i \cdot \widehat j = 0$, $\widehat j \cdot \widehat i = 0$, and $\widehat i \cdot \widehat i = 1$, $\widehat j \cdot \widehat j = 1$: $\left( {\overrightarrow a \times \widehat k} \right)^2 = (-x)^2 (\widehat j \cdot \widehat j) + 0 + 0 + (y)^2 (\widehat i \cdot \widehat i)$ $\left( {\overrightarrow a \times \widehat k} \right)^2 = x^2 (1) + y^2 (1)$ $\left( {\overrightarrow a \times \widehat k} \right)^2 = x^2 + y^2$

Step 7: Sum the squared magnitudes. Now, we add the results from Steps 2, 4, and 6: ${\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2} = (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2)$ $= y^2 + z^2 + x^2 + z^2 + x^2 + y^2$ $= 2x^2 + 2y^2 + 2z^2$ $= 2(x^2 + y^2 + z^2)$

Step 8: Relate the result to the magnitude of $\overrightarrow a$. Recall that the square of the magnitude of $\overrightarrow a$ is $|\overrightarrow a|^2 = x^2 + y^2 + z^2$. Therefore, the sum is: $2(x^2 + y^2 + z^2) = 2|\overrightarrow a|^2$ We can also write $|\overrightarrow a|^2$ as $\overrightarrow a^2$. $2|\overrightarrow a|^2 = 2\overrightarrow a^2$

Common Mistakes & Tips

• Sign Errors in Cross Products: Be very careful with the order of vectors in the cross product and the resulting signs ($\widehat i \times \widehat j = \widehat k$, but $\widehat j \times \widehat i = -\widehat k$).
• Dot Product of Unit Vectors: Remember that the dot product of orthogonal unit vectors is zero ($\widehat i \cdot \widehat j = 0$), while the dot product of a unit vector with itself is one ($\widehat i \cdot \widehat i = 1$).
• Magnitude Squared: The square of the magnitude of a vector is found by taking its dot product with itself, not by squaring each component individually and then summing.

Summary

The problem asks for the sum of the squares of the magnitudes of the cross products of a general vector $\overrightarrow a$ with the standard unit vectors $\widehat i$, $\widehat j$, and $\widehat k$. By expressing $\overrightarrow a$ in component form and systematically calculating each cross product and its squared magnitude, we found that the sum simplifies to $2(x^2 + y^2 + z^2)$. Recognizing that $x^2 + y^2 + z^2$ is the square of the magnitude of $\overrightarrow a$, i.e., $|\overrightarrow a|^2$ or $\overrightarrow a^2$, the final result is $2|\overrightarrow a|^2$.

The final answer is \boxed{2{\overrightarrow a ^2}} which corresponds to option (C).