Question:
Let \(\vec{A}=(\hat{i}+\hat{j})\) and,
\(\vec{B}=(2\hat{i}-\hat{j})\).
The magnitude of a coplanar vector \(\vec{C}\) such that
\(\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}=\vec{A}\cdot\vec{B}\),
is given by :
Let \(\vec{A}=(\hat{i}+\hat{j})\) and,
\(\vec{B}=(2\hat{i}-\hat{j})\).
The magnitude of a coplanar vector \(\vec{C}\) such that
\(\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}=\vec{A}\cdot\vec{B}\),
is given by :
\(\vec{B}=(2\hat{i}-\hat{j})\).
The magnitude of a coplanar vector \(\vec{C}\) such that
\(\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}=\vec{A}\cdot\vec{B}\),
is given by :
Show Hint
Compute \(\vec A\cdot\vec B\) first, write \(\vec C=a\hat i+b\hat j\) and solve the two dot-product equations.
Updated On: Nov 11, 2026
- \(\sqrt{\frac{10}{9}}\)
- \(\sqrt{\frac{5}{9}}\)
- \(\sqrt{\frac{20}{9}}\)
- \(\sqrt{\frac{9}{12}}\)
Show Solution
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The Correct Option is B
Solution and Explanation
\(\vec A\cdot\vec B=(1)(2)+(1)(-1)=1\).
Let \(\vec C=a\hat i+b\hat j\). \(\vec A\cdot\vec C=a+b=1\) and \(\vec B\cdot\vec C=2a-b=1\).
Adding: \(3a=2\Rightarrow a=\frac23\), then \(b=\frac13\).
\(|\vec C|=\sqrt{\frac49+\frac19}=\sqrt{\frac59}\).
Answer: (2) \(\sqrt{\frac{5}{9}}\)
Let \(\vec C=a\hat i+b\hat j\). \(\vec A\cdot\vec C=a+b=1\) and \(\vec B\cdot\vec C=2a-b=1\).
Adding: \(3a=2\Rightarrow a=\frac23\), then \(b=\frac13\).
\(|\vec C|=\sqrt{\frac49+\frac19}=\sqrt{\frac59}\).
Answer: (2) \(\sqrt{\frac{5}{9}}\)
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