Question

If $\vec{a}=3\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+7\hat{k}$ and $\vec{c}=7\hat{i}-\hat{j}+23\hat{k}$ are three vectors, then which of following statements is true ?

• A. a, b, c are mutually perpendicular
• B. a, b and c are coplanar
• C. a, b and c are non-coplanar
• D. a and b are collinear

Answer 1

Answer: (C)

a, b and c are non-coplanar

Answer 2

To determine the correct statement, we analyze each option:

1. Check for perpendicularity: We compute the dot product of $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{b} = (3)(2) + (1)(-1) + (-1)(7) = 6 - 1 - 7 = -2 \neq 0$. Thus, $\vec{a}$ and $\vec{b}$ are not perpendicular, so the vectors are not mutually perpendicular.

2. Check for coplanarity: We compute the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$. First, we find the cross product $\vec{b} \times \vec{c}$:

   $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 7 \\ 7 & -1 & 23 \end{vmatrix} = \hat{i}((-1)(23) - (7)(-1)) - \hat{j}((2)(23) - (7)(7)) + \hat{k}((2)(-1) - (-1)(7)) = \hat{i}(-23 + 7) - \hat{j}(46 - 49) + \hat{k}(-2 + 7) = -16\hat{i} + 3\hat{j} + 5\hat{k}$

   Now, we compute the dot product of $\vec{a}$ and $(\vec{b} \times \vec{c})$:

   $\vec{a} \cdot (\vec{b} \times \vec{c}) = (3)(-16) + (1)(3) + (-1)(5) = -48 + 3 - 5 = -50 \neq 0$.

   Since the scalar triple product is non-zero, the vectors are non-coplanar.

3. Check for collinearity: For $\vec{a}$ and $\vec{b}$ to be collinear, one must be a scalar multiple of the other. $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + 7\hat{k}$. There is no scalar $k$ such that $\vec{a} = k\vec{b}$, so they are not collinear.

Therefore, the correct statement is that a, b, and c are non-coplanar.