MHT CET · Maths · Vector Algebra
If the vectors \(\bar{a}=\hat{\imath}-2 \hat{\jmath}+\hat{k}, \bar{b}=2 \hat{\imath}-5 \hat{\jmath}+p \hat{k}\) and \(\bar{c}=5 \hat{\imath}-9 \hat{\jmath}+4 \hat{k}\) are coplanar,
then the value of \(\mathrm{p}\) is
- A \(-3\)
- B 3
- C \(\frac{1}{3}\)
- D \(-\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
Given vectors are coplanar \(\Rightarrow \bar{a} \cdot(\bar{b} \times \bar{c})=0\)
\(\begin{aligned}
&\left|\begin{array}{rrr}
1 & -2 & 1 \\
2 & -5 & \mathrm{P} \\
5 & -9 & 4
\end{array}\right|=0\end{aligned}\)
\(\therefore 1(-20+9 \mathrm{P})+2(8-5 \mathrm{P})+1(-18+25)=0\)
\(-20+9 \mathrm{P}+16-10 \mathrm{P}+7=0 \Rightarrow-\mathrm{P}+3=0 \Rightarrow\mathrm{P}=3\)
\(\begin{aligned}
&\left|\begin{array}{rrr}
1 & -2 & 1 \\
2 & -5 & \mathrm{P} \\
5 & -9 & 4
\end{array}\right|=0\end{aligned}\)
\(\therefore 1(-20+9 \mathrm{P})+2(8-5 \mathrm{P})+1(-18+25)=0\)
\(-20+9 \mathrm{P}+16-10 \mathrm{P}+7=0 \Rightarrow-\mathrm{P}+3=0 \Rightarrow\mathrm{P}=3\)
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