MHT CET · Maths · Vector Algebra
\(\overline{\mathrm{r}}=-4 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) is a linear combination of the vector \(\bar{a}=-\hat{i}-4 \hat{j}+3 \hat{k}\) and \(\bar{b}=-8 \hat{i}-\hat{j}+3 \hat{k}\), then
- A \(\overline{\mathrm{r}}=\frac{-4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}\)
- B \(\overline{\mathrm{r}}=\frac{4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}\)
- C \(\overline{\mathrm{r}}=\frac{-1}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}\)
- D \(\overline{\mathrm{r}}=\frac{1}{3} \overline{\mathrm{a}}-\frac{1}{3} \overline{\mathrm{b}}\)
Answer & Solution
Correct Answer
(A) \(\overline{\mathrm{r}}=\frac{-4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}\)
Step-by-step Solution
Detailed explanation
Let \(r=x \bar{a}+y \bar{b}\)
\(
\begin{aligned}
& \therefore-4 \hat{i}-6 \hat{j}-2 \hat{k}=x(-\hat{i}+4 \hat{j}+3 \hat{k})+y(-8 \hat{i}-\hat{j}+3 \hat{k}) \\
& =(-x-8 y) \hat{i}+(4 x-y) \hat{j}+(3 x+3 y) \hat{k} \\
& \therefore \quad-x-8 y \quad \ldots(1) \\
& 4 x-y \quad=-4 \quad \ldots(2) \\
& 3 x+3 y \quad=-2 \quad \ldots(3)
\end{aligned}
\)
Solving (1) and (2), we get \(\mathrm{y}=\frac{2}{3}\) and \(\mathrm{x}=\frac{-4}{3}\)
\(
\begin{aligned}
& \therefore-4 \hat{i}-6 \hat{j}-2 \hat{k}=x(-\hat{i}+4 \hat{j}+3 \hat{k})+y(-8 \hat{i}-\hat{j}+3 \hat{k}) \\
& =(-x-8 y) \hat{i}+(4 x-y) \hat{j}+(3 x+3 y) \hat{k} \\
& \therefore \quad-x-8 y \quad \ldots(1) \\
& 4 x-y \quad=-4 \quad \ldots(2) \\
& 3 x+3 y \quad=-2 \quad \ldots(3)
\end{aligned}
\)
Solving (1) and (2), we get \(\mathrm{y}=\frac{2}{3}\) and \(\mathrm{x}=\frac{-4}{3}\)
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