Question

If \(\quad \overline{\mathrm{a}}=\lambda x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+4 z \hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=y \hat{\mathrm{i}}+x \hat{\mathrm{j}}+3 y \hat{\mathrm{k}}\), \(\overline{\mathrm{c}}=-z \hat{\mathrm{i}}-2 z \hat{\mathrm{j}}-(\lambda+1) \hat{\mathrm{k}} x\) are the sides of the triangle ABC , where \(x, y, \mathrm{z}\) are not all zero, such that \(\bar{a}+\bar{b}-\bar{c}=\overline{0}\), then value of \(\lambda\) is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Equate each component of the vector sum to zero and use the condition for a non-trivial solution of the homogeneous system.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The vector equation \(\bar{a}+\bar{b}-\bar{c}=\bar{0}\) gives three linear equations in \(x, y, z\). Since \(x, y, z\) are not all zero, the determinant of the coefficients must be zero.

Step 2: Form the equations
Equating the \(\hat{i}, \hat{j}, \hat{k}\) components:
\(\lambda x+y+z=0\)
\(x+y+2z=0\)
\((\lambda+1)x+3y+4z=0\)

Step 3: Calculation
Determinant: \(\begin{vmatrix}\lambda&1&1\\1&1&2\\\lambda+1&3&4\end{vmatrix}=\lambda(4-6)-1(4-2\lambda-2)+1(3-\lambda-1)\)
\(=-2\lambda-2+2\lambda+2-\lambda=-\lambda\)

Step 4: Conclusion
Setting the determinant equal to zero gives \(\lambda=0\).

Final Answer:
The value of \(\lambda\) is 0. \[ \boxed{\text{(A) }0} \]