Question

If \( \vec{a} = 2\vec{i} - \vec{j} + 6\vec{k} \); \( \vec{b} = \vec{i} - \vec{j} + \vec{k} \) and \( \vec{c} = 3\vec{j} - \vec{k} \), then \( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \)

• A. \( 20\vec{i} + 3\vec{j} - 4\vec{k} \)
• B. \( 20\vec{i} - 3\vec{j} + 4\vec{k} \)
• C. \( 3\vec{i} + 20\vec{j} - 4\vec{k} \)
• D. \( 4\vec{i} + 20\vec{j} - 3\vec{k} \) Correct Answer

Hint:

The cross product \( \vec{A} \times \vec{B} \) can be calculated using the determinant form: \( \vec{A} \times \vec{B} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}
A_x & A_y & A_z
B_x & B_y & B_z \end{vmatrix} \). Be careful with signs during calculation. The expression \( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \) is also equal to \( (\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}) \), which represents twice the vector area of the triangle with vertices \( \vec{a}, \vec{b}, \vec{c} \) (if these are position vectors).

Answer 1

The Correct Option is A

This expression \( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \) is related to the area of a triangle with vertices at positions given by \( \vec{a}, \vec{b}, \vec{c} \) if they are position vectors.
However, here they are just vectors.
A property is that if \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) (forming a triangle), then \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \).
In this case, the sum is \( 3(\vec{a} \times \vec{b}) \).
But this condition is not given.
We need to calculate each cross product.

Step 1: Calculate \( \vec{a} \times \vec{b} \).
\( \vec{a} = (2, -1, 6) \), \( \vec{b} = (1, -1, 1) \) \[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 6 \1 & -\1 & 1 \end{vmatrix} = \vec{i}((-1)(1) - (6)(-1)) - \vec{j}((2)(1) - (6)(1)) + \vec{k}((2)(-1) - (-1)(1)) \] \[ = \vec{i}(-1+6) - \vec{j}(2-6) + \vec{k}(-2+1) = 5\vec{i} - (-4)\vec{j} + (-1)\vec{k} = 5\vec{i} + 4\vec{j} - \vec{k} \]
Step 2: Calculate \( \vec{b} \times \vec{c} \).
\( \vec{b} = (1, -1, 1) \), \( \vec{c} = (0, 3, -1) \) \[ \vec{b} \times \vec{c} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 1 &\ -1 & 1 \ 0 &\ 3 & -1 \end{vmatrix} = \vec{i}((-1)(-1) - (1)(3)) - \vec{j}((1)(-1) - (1)(0)) + \vec{k}((1)(3) - (-1)(0)) \] \[ = \vec{i}(1-3) - \vec{j}(-1-0) + \vec{k}(3-0) = -2\vec{i} - (-1)\vec{j} + 3\vec{k} = -2\vec{i} + \vec{j} + 3\vec{k} \]
Step 3: Calculate \( \vec{c} \times \vec{a} \).
\( \vec{c} = (0, 3, -1) \), \( \vec{a} = (2, -1, 6) \) \[ \vec{c} \times \vec{a} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 3 & -1 \\ 2 & -1 & 6 \end{vmatrix} = \vec{i}((3)(6) - (-1)(-1)) - \vec{j}((0)(6) - (-1)(2)) + \vec{k}((0)(-1) - (3)(2)) \] \[ = \vec{i}(18-1) - \vec{j}(0+2) + \vec{k}(0-6) = 17\vec{i} - 2\vec{j} - 6\vec{k} \]
Step 4: Sum the three cross products.
\( (5\vec{i} + 4\vec{j} - \vec{k}) + (-2\vec{i} + \vec{j} + 3\vec{k}) + (17\vec{i} - 2\vec{j} - 6\vec{k}) \) Combine \( \vec{i} \) components: \( 5 - 2 + 17 = 3 + 17 = 20 \).
Combine \( \vec{j} \) components: \( 4 + 1 - 2 = 5 - 2 = 3 \).
Combine \( \vec{k} \) components: \( -1 + 3 - 6 = 2 - 6 = -4 \).
Result: \( 20\vec{i} + 3\vec{j} - 4\vec{k} \).
This matches option (1).