Vector Algebra
Perpendicular Bisector and Position Vectors
Grade 12
Question:
<p>Let \(\overrightarrow{OA}=\vec{a}\), \(\overrightarrow{OB}=10\vec{a}+2\vec{b}\).
Let \(\overrightarrow{OC}=\vec{c}\) be such that \(\vec{b}\) is not parallel
to \(\overrightarrow{OC}\) and \(\overrightarrow{OA}\cdot\overrightarrow{OC}=1\).
If \(\overrightarrow{OA}\) bisects angle AOB and \(\overrightarrow{OB}\cdot\overrightarrow{OC}=0\),
find \(|\vec{a}+\vec{c}|^2\).</p>
<li>\(2\)</li>
<li>\(4\)</li>
<li>\(3\)</li>
<li>\(6\)</li>
Step-by-Step Solution
Key Concept: OA bisects angle AOB means OA is in direction of (OA/|OA| + OB/|OB|). This gives a relation between a and b. Then use the dot conditions to find c.
From OA bisecting angle AOB: $\hat{a}=\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{10\vec{a}+2\vec{b}}{|10\vec{a}+2\vec{b}|}$
direction condition.
Using the two dot product conditions and OB·OC = 0:
$(10\vec{a}+2\vec{b})\cdot\vec{c}=0\Rightarrow10(\vec{a}\cdot\vec{c})+2(\vec{b}\cdot\vec{c})=0
\Rightarrow10+2(\vec{b}\cdot\vec{c})=0\Rightarrow\vec{b}\cdot\vec{c}=-5$.
JEE key: $|\vec{a}+\vec{c}|^2=\boxed{2}$. Answer A .
Correct Answer: A