MHT CET · Maths · Vector Algebra
Let \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\), and \(\overline{\mathrm{c}}\) be three non-zero vectors such that no two of these are collinear. If the vector \(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}\) is collinear with \(\overline{\mathrm{c}}\) and \(\overline{\mathrm{b}}+3 \overline{\mathrm{c}}\) is collinear with \(\overline{\mathrm{a}}\), then \(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}\) equals
- A \(\lambda \bar{c}\) ( \(\lambda\) being some non-zero scalar)
- B \(\lambda \bar{b}\) ( \(\lambda\) being some non-zero scalar)
- C \(\lambda \bar{a}\) ( \(\lambda\) being some non-zero scalar)
- D \(\overline{0}\) ( \(\lambda\) being some non-zero scalar)
Answer & Solution
Correct Answer
(D) \(\overline{0}\) ( \(\lambda\) being some non-zero scalar)
Step-by-step Solution
Detailed explanation
\(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}\) is collinear with \(\overline{\mathrm{c}}\)
\(\therefore \quad \bar{a}+2 \bar{b}=n \bar{c}\)...(i)
Similarly \(\overline{\mathrm{b}}+3 \overline{\mathrm{c}}=\mathrm{m} \overline{\mathrm{a}}\)...(ii)
\(m\) and \(n\) are non-zero scalars.
\(\therefore \quad\) (i) \(\Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}=(\mathrm{n}+6) \overline{\mathrm{c}}\)
(ii) \(\Rightarrow \bar{a}+2 \bar{b}+6 \bar{c}=(2 m+1) \bar{a}\)
\(\Rightarrow \mathrm{n}+6=0\) and \(2 \mathrm{~m}+1=0\)
\(\Rightarrow \mathrm{n}=-6\) and \(\mathrm{m}=\frac{-1}{2}\)
\(\therefore \quad\) (i) \(\Rightarrow \bar{a}+2 \bar{b}+6 \bar{c}=0\)
\(\therefore \quad \bar{a}+2 \bar{b}=n \bar{c}\)...(i)
Similarly \(\overline{\mathrm{b}}+3 \overline{\mathrm{c}}=\mathrm{m} \overline{\mathrm{a}}\)...(ii)
\(m\) and \(n\) are non-zero scalars.
\(\therefore \quad\) (i) \(\Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}=(\mathrm{n}+6) \overline{\mathrm{c}}\)
(ii) \(\Rightarrow \bar{a}+2 \bar{b}+6 \bar{c}=(2 m+1) \bar{a}\)
\(\Rightarrow \mathrm{n}+6=0\) and \(2 \mathrm{~m}+1=0\)
\(\Rightarrow \mathrm{n}=-6\) and \(\mathrm{m}=\frac{-1}{2}\)
\(\therefore \quad\) (i) \(\Rightarrow \bar{a}+2 \bar{b}+6 \bar{c}=0\)
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