Key Concepts and Formulas

• Vector Addition and Subtraction: To add or subtract vectors, we add or subtract their corresponding components. If $\overrightarrow u = u_1\widehat i + u_2\widehat j + u_3\widehat k$ and $\overrightarrow v = v_1\widehat i + v_2\widehat j + v_3\widehat k$, then $\overrightarrow u + \overrightarrow v = (u_1+v_1)\widehat i + (u_2+v_2)\widehat j + (u_3+v_3)\widehat k$ and $\overrightarrow u - \overrightarrow v = (u_1-v_1)\widehat i + (u_2-v_2)\widehat j + (u_3-v_3)\widehat k$.
• Cross Product: A vector perpendicular to two given vectors $\overrightarrow u$ and $\overrightarrow v$ can be found using their cross product, $\overrightarrow u \times \overrightarrow v$. The magnitude of the cross product is $||\overrightarrow u \times \overrightarrow v|| = ||\overrightarrow u|| ||\overrightarrow v|| \sin\theta$, where $\theta$ is the angle between $\overrightarrow u$ and $\overrightarrow v$. The direction of $\overrightarrow u \times \overrightarrow v$ is perpendicular to the plane containing $\overrightarrow u$ and $\overrightarrow v$, following the right-hand rule. The cross product can be calculated using a determinant: $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$
• Magnitude of a Vector: The magnitude of a vector $\overrightarrow v = x\widehat i + y\widehat j + z\widehat k$ is given by $||\overrightarrow v|| = \sqrt{x^2 + y^2 + z^2}$.
• Scalar Multiplication of a Vector: Multiplying a vector by a scalar $k$ scales its magnitude by $|k|$ and keeps its direction the same if $k > 0$, or reverses its direction if $k < 0$. If $\overrightarrow v = x\widehat i + y\widehat j + z\widehat k$, then $k\overrightarrow v = (kx)\widehat i + (ky)\widehat j + (kz)\widehat k$.

Step-by-Step Solution

1. Calculate the Vectors $(\overrightarrow a + \overrightarrow b)$ and $(\overrightarrow a - \overrightarrow b)$ We are given two vectors $\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k$ and $\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k$. We need to find a vector perpendicular to the sum and difference of these two vectors. First, let's compute $\overrightarrow a + \overrightarrow b$: $\overrightarrow a + \overrightarrow b = (3\widehat i + 2\widehat j + 2\widehat k) + (\widehat i + 2\widehat j - 2\widehat k)$ $\overrightarrow a + \overrightarrow b = (3+1)\widehat i + (2+2)\widehat j + (2-2)\widehat k$ $\overrightarrow a + \overrightarrow b = 4\widehat i + 4\widehat j + 0\widehat k = 4\widehat i + 4\widehat j$ Next, let's compute $\overrightarrow a - \overrightarrow b$: $\overrightarrow a - \overrightarrow b = (3\widehat i + 2\widehat j + 2\widehat k) - (\widehat i + 2\widehat j - 2\widehat k)$ $\overrightarrow a - \overrightarrow b = (3-1)\widehat i + (2-2)\widehat j + (2-(-2))\widehat k$ $\overrightarrow a - \overrightarrow b = 2\widehat i + 0\widehat j + 4\widehat k = 2\widehat i + 4\widehat k$ Why this step? The problem requires us to find a vector perpendicular to specific combinations of $\overrightarrow a$ and $\overrightarrow b$. Calculating these combinations first provides the two vectors for which we need to find a common perpendicular.

2. Find a Vector Perpendicular to $(\overrightarrow a + \overrightarrow b)$ and $(\overrightarrow a - \overrightarrow b)$ using the Cross Product Let $\overrightarrow u = \overrightarrow a + \overrightarrow b = 4\widehat i + 4\widehat j$ and $\overrightarrow v = \overrightarrow a - \overrightarrow b = 2\widehat i + 4\widehat k$. A vector perpendicular to both $\overrightarrow u$ and $\overrightarrow v$ is given by their cross product $\overrightarrow u \times \overrightarrow v$. $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$ Expanding the determinant: $\overrightarrow u \times \overrightarrow v = \widehat i ((4)(4) - (0)(0)) - \widehat j ((4)(4) - (0)(2)) + \widehat k ((4)(0) - (4)(2))$ $\overrightarrow u \times \overrightarrow v = \widehat i (16 - 0) - \widehat j (16 - 0) + \widehat k (0 - 8)$ $\overrightarrow u \times \overrightarrow v = 16\widehat i - 16\widehat j - 8\widehat k$ Let this perpendicular vector be $\overrightarrow P = 16\widehat i - 16\widehat j - 8\widehat k$. We can factor out a common term to simplify the vector for further calculations. The greatest common divisor of the components is 8. $\overrightarrow P = 8(2\widehat i - 2\widehat j - \widehat k)$ Why this step? The cross product is the standard method to find a vector orthogonal to two given vectors. Factoring out a common scalar simplifies the vector's components, making the magnitude calculation easier and aligning it with the form of the answer options.

3. Scale the Perpendicular Vector to the Desired Magnitude We need a vector with a magnitude of 12. The vector $\overrightarrow P = 16\widehat i - 16\widehat j - 8\widehat k$ is in the correct direction. Let the required vector be $\overrightarrow r$. Then $\overrightarrow r$ must be a scalar multiple of $\overrightarrow P$. $\overrightarrow r = k \overrightarrow P = k (16\widehat i - 16\widehat j - 8\widehat k)$ We are given that $||\overrightarrow r|| = 12$. Let's calculate the magnitude of $\overrightarrow P$: $||\overrightarrow P|| = ||16\widehat i - 16\widehat j - 8\widehat k|| = \sqrt{16^2 + (-16)^2 + (-8)^2}$ $||\overrightarrow P|| = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$ Now, we relate the magnitudes: $||\overrightarrow r|| = ||k \overrightarrow P|| = |k| ||\overrightarrow P||$ $12 = |k| \cdot 24$ Solving for $|k|$: $|k| = \frac{12}{24} = \frac{1}{2}$ This means $k = \frac{1}{2}$ or $k = -\frac{1}{2}$. Why this step? The cross product gives a vector of the correct direction but not necessarily the correct magnitude. Scalar multiplication is used to adjust the magnitude to the required value. Considering both positive and negative values for the scalar $k$ accounts for the two possible directions of a vector with a given magnitude.

4. Construct the Final Vector We have two possible values for $k$: $\frac{1}{2}$ and $-\frac{1}{2}$.

Case 1: $k = \frac{1}{2}$ $\overrightarrow r = \frac{1}{2} (16\widehat i - 16\widehat j - 8\widehat k) = 8\widehat i - 8\widehat j - 4\widehat k$ We can factor out 4 from this vector: $\overrightarrow r = 4(2\widehat i - 2\widehat j - \widehat k)$

Case 2: $k = -\frac{1}{2}$ $\overrightarrow r = -\frac{1}{2} (16\widehat i - 16\widehat j - 8\widehat k) = -8\widehat i + 8\widehat j + 4\widehat k$ We can factor out -4 from this vector: $\overrightarrow r = -4(2\widehat i - 2\widehat j - \widehat k)$ Now, let's compare these results with the given options. Option (A) is $4(2\widehat i - 2\widehat j - \widehat k)$. This matches our result from Case 1.

Why this step? This step involves substituting the determined scalar multiplier(s) back into the vector expression to obtain the final answer(s), ensuring both the correct direction and magnitude are achieved.

Common Mistakes & Tips

• Order of Cross Product: The order of vectors in the cross product matters: $\overrightarrow u \times \overrightarrow v = -(\overrightarrow v \times \overrightarrow u)$. However, both $\overrightarrow u \times \overrightarrow v$ and $\overrightarrow v \times \overrightarrow u$ are perpendicular to both $\overrightarrow u$ and $\overrightarrow v$. Since the question asks for "one such vector," either direction is valid if it leads to one of the options.
• Magnitude Calculation: Be careful when squaring negative components during magnitude calculations (e.g., $(-16)^2 = 256$).
• Scalar Multiplier Sign: When scaling a vector to a specific magnitude, remember that the scalar multiplier can be positive or negative, leading to two possible vectors pointing in opposite directions but having the same magnitude.

Summary

To find a vector perpendicular to two given vectors with a specific magnitude, we first compute the two vectors in question, $(\overrightarrow a + \overrightarrow b)$ and $(\overrightarrow a - \overrightarrow b)$. Then, we find a vector perpendicular to these two using their cross product. This cross product gives us the direction. Finally, we scale this direction vector by a scalar factor to achieve the desired magnitude of 12. This involves calculating the magnitude of the cross product vector and then determining the appropriate scalar multiplier, which can be positive or negative.

The final answer is $\boxed{\text{4}\left( {2\widehat i - 2\widehat j - \widehat k} \right)}$.