Question:
In \(△OAB\), \(O(0,0,0), A(6,2,-3)\) and \(B(4,0,3)\) are the vertices. Let \(\overset{⃗}{a}\) and \(\overset{⃗}{b}\) be position vectors of points \(A\) and \(B\) respectively and \(OM\) is the projection of \(\overset{⃗}{a}\) on \(\overset{⃗}{b}\) then \(l(AM)\) is equal to...
In \(△OAB\), \(O(0,0,0), A(6,2,-3)\) and \(B(4,0,3)\) are the vertices. Let \(\overset{⃗}{a}\) and \(\overset{⃗}{b}\) be position vectors of points \(A\) and \(B\) respectively and \(OM\) is the projection of \(\overset{⃗}{a}\) on \(\overset{⃗}{b}\) then \(l(AM)\) is equal to...
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\(AM\) is the perpendicular from \(A\) to line \(OB\).
Updated On: Oct 1, 2026
- \(\sqrt{10} units\)
- \(2\sqrt{10} units\)
- \(10 units\)
- \(40 units\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
\(OM\) is the component of \(\vec a\) along \(\vec b\), so \(M\) is the foot of the perpendicular from \(A\) to \(OB\). Then \(AM\) is the perpendicular distance.
Step 2: Key Formula or Approach
\(OM=\dfrac{\vec a\cdot\vec b}{|\vec b|}\) and \(AM^2=|\vec a|^2-OM^2\).
Step 3: Detailed Explanation
\(\vec a\cdot\vec b=24+0-9=15\) and \(|\vec b|=\sqrt{16+0+9}=5\), so \(OM=3\).
\(|\vec a|=\sqrt{36+4+9}=7\).
\[ AM^2=49-9=40 \Rightarrow AM=2\sqrt{10} \]
Final Answer:
The length \(AM=2\sqrt{10}\) units, option (B). \[ \boxed{2\sqrt{10}\ \text{units (B)}} \]
\(OM\) is the component of \(\vec a\) along \(\vec b\), so \(M\) is the foot of the perpendicular from \(A\) to \(OB\). Then \(AM\) is the perpendicular distance.
Step 2: Key Formula or Approach
\(OM=\dfrac{\vec a\cdot\vec b}{|\vec b|}\) and \(AM^2=|\vec a|^2-OM^2\).
Step 3: Detailed Explanation
\(\vec a\cdot\vec b=24+0-9=15\) and \(|\vec b|=\sqrt{16+0+9}=5\), so \(OM=3\).
\(|\vec a|=\sqrt{36+4+9}=7\).
\[ AM^2=49-9=40 \Rightarrow AM=2\sqrt{10} \]
Final Answer:
The length \(AM=2\sqrt{10}\) units, option (B). \[ \boxed{2\sqrt{10}\ \text{units (B)}} \]
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