MHT CET · Maths · Vector Algebra
If \(\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{k}}, \overline{\mathrm{b}}=x \hat{\mathrm{i}}+\hat{\mathrm{j}}+(1-x) \hat{\mathrm{k}} \quad\) and \(\overline{\mathrm{c}}=y \hat{\mathrm{i}}+x \hat{\mathrm{j}}+(1+x-y) \hat{\mathrm{k}}\) then \(\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})\) depends on
- A only \(x\)
- B only \(y\)
- C neither \(x\) nor \(y\)
- D both \(x\) and \(y\)
Answer & Solution
Correct Answer
(C) neither \(x\) nor \(y\)
Step-by-step Solution
Detailed explanation
\(\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|\)
Applying \(\mathrm{C}_3 \Rightarrow \mathrm{C}_3+\mathrm{C}_1\), we get
\([\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]=\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & 1 & 1 \\
y & x & 1+x
\end{array}\right|=1(1+x-x)=1\)
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|\)
Applying \(\mathrm{C}_3 \Rightarrow \mathrm{C}_3+\mathrm{C}_1\), we get
\([\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]=\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & 1 & 1 \\
y & x & 1+x
\end{array}\right|=1(1+x-x)=1\)
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