Question:
If the vectors \( 4\hat{i} + 11\hat{j} + m\hat{k},\ 7\hat{i} + 2\hat{j} + 6\hat{k} \) and \( \hat{i} + 5\hat{j} + 4\hat{k} \) are coplanar, then \( m \) is equal to
If the vectors \( 4\hat{i} + 11\hat{j} + m\hat{k},\ 7\hat{i} + 2\hat{j} + 6\hat{k} \) and \( \hat{i} + 5\hat{j} + 4\hat{k} \) are coplanar, then \( m \) is equal to
Show Hint
Coplanarity always reduces to determinant = 0.
Updated On: May 1, 2026
- \( 38 \)
- \( 0 \)
- \( 10 \)
- \( -10 \)
- \( 25 \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
Coplanar vectors ⇒ scalar triple product = 0.
Step 1: Form determinant.
\[ \begin{vmatrix} 4 & 11 & m 7 & 2 & 6 1 & 5 & 4 \end{vmatrix} = 0 \]
Step 2: Expand determinant.
\[ = 4(2\cdot4 - 6\cdot5) - 11(7\cdot4 - 6\cdot1) + m(7\cdot5 - 2\cdot1) \]
Step 3: Simplify each term.
\[ 4(8-30) -11(28-6) + m(35-2) \] \[ = 4(-22) -11(22) + 33m \]
Step 4: Compute values.
\[ -88 -242 + 33m = 0 \]
Step 5: Solve equation.
\[ 33m = 330 \Rightarrow m = 10 \]
Step 1: Form determinant.
\[ \begin{vmatrix} 4 & 11 & m 7 & 2 & 6 1 & 5 & 4 \end{vmatrix} = 0 \]
Step 2: Expand determinant.
\[ = 4(2\cdot4 - 6\cdot5) - 11(7\cdot4 - 6\cdot1) + m(7\cdot5 - 2\cdot1) \]
Step 3: Simplify each term.
\[ 4(8-30) -11(28-6) + m(35-2) \] \[ = 4(-22) -11(22) + 33m \]
Step 4: Compute values.
\[ -88 -242 + 33m = 0 \]
Step 5: Solve equation.
\[ 33m = 330 \Rightarrow m = 10 \]
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