Question

Let \(\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}\), \(\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}\) and \(\vec{c}\) be a vector such that \(\vec{a} \times \vec{c} = \vec{b}\). If \(\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k}) = 4\), then \(|\vec{a}+\vec{c}|^2\) is equal to:

• A. 33
• B. 35
• C. 27
• D. 30

Hint:

When given a vector equation like \(\vec{a} \times \vec{c} = \vec{b}\), a direct approach is to write \(\vec{c}\) in component form and solve the resulting system of linear equations.
Always check for consistency and dependence in the equations. In this case, three equations were derived from the cross product, but one was redundant, requiring the use of the fourth condition from the dot product to find a unique solution.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two vectors \(\vec{a}\) and \(\vec{b}\), and a third vector \(\vec{c}\) which is defined by two conditions: one involving a cross product with \(\vec{a}\) and the other involving a dot product. We need to find the squared magnitude of the vector sum \(\vec{a}+\vec{c}\).
Step 2: Key Formula or Approach:
1. Let \(\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}\). 2. Use the given conditions to form a system of linear equations for the components \(c_1, c_2, c_3\). 3. The cross product \(\vec{a} \times \vec{c}\) can be computed using the determinant formula. 4. The dot product \(\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k})\) gives a simple linear equation. 5. Solve the system of equations to find \(\vec{c}\). 6. Calculate \(\vec{a}+\vec{c}\) and then its squared magnitude.
Step 3: Detailed Explanation:
Let \(\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}\).
From the first condition, \(\vec{a} \times \vec{c} = \vec{b}\):
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ c_1 & c_2 & c_3 \end{vmatrix} = \hat{i}(2c_3 - 2c_2) - \hat{j}(-c_3 - 2c_1) + \hat{k}(-c_2 - 2c_1) \] \[ (2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k} = 8\hat{i} + 7\hat{j} - 3\hat{k} \] Comparing the components, we get a system of three linear equations: 1. \(2c_3 - 2c_2 = 8 \implies c_3 - c_2 = 4\) 2. \(c_3 + 2c_1 = 7\) 3. \(-(c_2 + 2c_1) = -3 \implies c_2 + 2c_1 = 3\) From the second given condition, \(\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k}) = 4\):
\[ (c_1\hat{i} + c_2\hat{j} + c_3\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k}) = 4 \] 4. \(c_1 + c_2 + c_3 = 4\) Now we solve this system of equations. From (3), we get \(c_2 = 3 - 2c_1\). Substitute this into (1): \(c_3 - (3 - 2c_1) = 4 \implies c_3 + 2c_1 - 3 = 4 \implies c_3 + 2c_1 = 7\). This is the same as equation (2), which means the first three equations are dependent. We must use equation (4). Substitute \(c_2 = 3 - 2c_1\) into (1) to get \(c_3 = 4 + c_2 = 4 + (3 - 2c_1) = 7 - 2c_1\). Now substitute both \(c_2\) and \(c_3\) into equation (4): \[ c_1 + (3 - 2c_1) + (7 - 2c_1) = 4 \] \[ 10 - 3c_1 = 4 \] \[ 3c_1 = 6 \implies c_1 = 2 \] Now find \(c_2\) and \(c_3\): \[ c_2 = 3 - 2(2) = 3 - 4 = -1 \] \[ c_3 = 7 - 2(2) = 7 - 4 = 3 \] So, the vector \(\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}\). Now we find \(\vec{a}+\vec{c}\): \[ \vec{a}+\vec{c} = (-\hat{i} + 2\hat{j} + 2\hat{k}) + (2\hat{i} - \hat{j} + 3\hat{k}) = (2-1)\hat{i} + (2-1)\hat{j} + (2+3)\hat{k} = \hat{i} + \hat{j} + 5\hat{k} \] Finally, we calculate its squared magnitude: \[ |\vec{a}+\vec{c}|^2 = 1^2 + 1^2 + 5^2 = 1 + 1 + 25 = 27 \] Step 4: Final Answer:
The value of \(|\vec{a}+\vec{c}|^2\) is 27.