MHT CET · Maths · Vector Algebra
\(\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}\) are three vectors such that \(\vec{a}=x \vec{b}+y \vec{c}\), then \(x+y=\)
- A -1
- B -2
- C 5
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(\vec{a}=x \vec{b}+y \vec{c} \)
\( \therefore 4 \hat{i}+13 \hat{j}-18 \hat{k}=(\hat{i}-2 \hat{j}+3 \hat{k})(x)+(2 \hat{i}+3 \hat{j}-4 \hat{k})(y) \)
\( =(x+2 y) \hat{i}+(-2 x+3 y) \hat{j}+(3 x-4 y) \hat{k} \)
\( \therefore x+2 y=4,-2 x+3 y=13 \text { and } 3 x-4 y=-18\)
Solving, we get \(x=-2, y=3 \Rightarrow x+y=1\)
\( \therefore 4 \hat{i}+13 \hat{j}-18 \hat{k}=(\hat{i}-2 \hat{j}+3 \hat{k})(x)+(2 \hat{i}+3 \hat{j}-4 \hat{k})(y) \)
\( =(x+2 y) \hat{i}+(-2 x+3 y) \hat{j}+(3 x-4 y) \hat{k} \)
\( \therefore x+2 y=4,-2 x+3 y=13 \text { and } 3 x-4 y=-18\)
Solving, we get \(x=-2, y=3 \Rightarrow x+y=1\)
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