MHT CET · Maths · Vector Algebra
\(\bar{a}=\hat{i}+j+\hat{k}, \bar{b}=\hat{i}-j+2 \hat{k}\) and \(\bar{c}=x|+(x-1)\rangle-\hat{k}\). If the vector \(\bar{c}\) lies in the plane
of \(\bar{u}\) and \(\bar{b}\), then \(x=\)
- A \(\frac{2}{3}\)
- B \(\frac{-3}{2}\)
- C \(\frac{-2}{3}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{-3}{2}\)
Step-by-step Solution
Detailed explanation
Given vectors are coplanar, we write : \(\bar{a} \cdot(\bar{b} \times \bar{c})=0\)
\(\begin{aligned}
&\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 2 \\
x & x-1 & -1
\end{array}\right|=0 \\
\therefore & 1(1-2 x+2)-1(-1-2 x)+1(x-1+x)=0 \\
& 3-2 x+1+2 x+2 x-1=0 \Rightarrow 2 x+3=0 \Rightarrow x=\frac{-3}{2}
\end{aligned}\)
\(\begin{aligned}
&\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 2 \\
x & x-1 & -1
\end{array}\right|=0 \\
\therefore & 1(1-2 x+2)-1(-1-2 x)+1(x-1+x)=0 \\
& 3-2 x+1+2 x+2 x-1=0 \Rightarrow 2 x+3=0 \Rightarrow x=\frac{-3}{2}
\end{aligned}\)
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