MHT CET · Maths · Three Dimensional Geometry
If the vector \(\overline{\mathrm{c}}\) lies in the plane of \(\overline{\mathrm{a}}\) and \(\bar{b}\), where \(\bar{a}=\hat{i}-\hat{j}+2 \hat{k}, \bar{b}=\hat{i}+\hat{j}+\hat{k}\) and \(\overline{\mathrm{c}}=x \hat{\mathrm{i}}-(2-x) \hat{\mathrm{j}}-\hat{\mathrm{k}}\), then the value of \(x\) is
- A 4
- B -4
- C 2
- D -2
Answer & Solution
Correct Answer
(D) -2
Step-by-step Solution
Detailed explanation
According to the given condition, vectors \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are coplanar.
\(\begin{aligned}
& \therefore \quad \overline{\mathrm{a}} \cdot(\overline{\mathrm{~b}} \times \overline{\mathrm{c}})=0 \\
& \therefore \quad\left|\begin{array}{ccc}
1 & -1 & 2 \\
1 & 1 & 1 \\
x & x-2 & -1
\end{array}\right|=0 \\
& \Rightarrow 1(-1-x+2)+1(-1-x)+2(x-2-x)=0 \\
& \\
& \Rightarrow 1-x-1-x-4=0 \\
& \Rightarrow x=-2
\end{aligned}\)
\(\begin{aligned}
& \therefore \quad \overline{\mathrm{a}} \cdot(\overline{\mathrm{~b}} \times \overline{\mathrm{c}})=0 \\
& \therefore \quad\left|\begin{array}{ccc}
1 & -1 & 2 \\
1 & 1 & 1 \\
x & x-2 & -1
\end{array}\right|=0 \\
& \Rightarrow 1(-1-x+2)+1(-1-x)+2(x-2-x)=0 \\
& \\
& \Rightarrow 1-x-1-x-4=0 \\
& \Rightarrow x=-2
\end{aligned}\)
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