Question

If the vectors \(2\hat{i}-\hat{j}+\hat{k}\), \(\hat{i}+2\hat{j}-3\hat{k}\) and \(3\hat{i}+a\hat{j}+5\hat{k}\) are coplanar, find the value of \(a\).

• A. \(4\)
• B. \(-4\)
• C. \(2\)
• D. \(-2\)

Hint:

Three vectors are coplanar if the determinant formed by their components equals zero. \[ \vec{A}\cdot(\vec{B}\times\vec{C})=0 \] This is a standard shortcut used in vector algebra problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem provides three vectors and states that they are coplanar.
We need to find the unknown scalar \(a\) that satisfies the condition for coplanarity.
Step 2: Key Formula or Approach:
For three vectors \(\vec{u}, \vec{v}, \text{ and } \vec{w}\) to be coplanar, their scalar triple product must be zero.
In terms of components, the determinant formed by the coefficients of \(\hat{i}, \hat{j}, \text{ and } \hat{k}\) must vanish:
\[ D = \begin{vmatrix} u_1 & u_2 & u_3
v_1 & v_2 & v_3
w_1 & w_2 & w_3 \end{vmatrix} = 0 \]
Step 3: Detailed Explanation:
Let the vectors be:
\(\vec{u} = 2\hat{i} - \hat{j} + \hat{k}\)
\(\vec{v} = \hat{i} + 2\hat{j} - 3\hat{k}\)
\(\vec{w} = 3\hat{i} + a\hat{j} + 5\hat{k}\)
Setting the determinant to zero:
\[ \begin{vmatrix} 2 & -1 & 1
1 & 2 & -3
3 & a & 5 \end{vmatrix} = 0 \]
Expanding the determinant along the first row:
\[ 2[(2 \times 5) - (-3 \times a)] - (-1)[(1 \times 5) - (-3 \times 3)] + 1[(1 \times a) - (2 \times 3)] = 0 \]
\[ 2[10 + 3a] + 1[5 + 9] + 1[a - 6] = 0 \]
\[ 20 + 6a + 14 + a - 6 = 0 \]
Combine the terms containing \(a\) and the constant terms:
\[ 7a + 28 = 0 \]
\[ 7a = -28 \]
\[ a = -4 \]
Step 4: Final Answer:
The value of \(a\) for which the vectors are coplanar is \(-4\).