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...

• A. \(\sqrt{10} units\)
• B. \(2\sqrt{10} units\)
• C. \(10 units\)
• D. \(40 units\)

Hint:

\(AM\) is the perpendicular from \(A\) to line \(OB\).

Answer 1

The Correct Option is B

Step 1: Use the cross product
$AM=\dfrac{|\vec a\times\vec b|}{|\vec b|}$.

Step 2: Compute
$\vec a\times\vec b=(2\cdot3-(-3)\cdot0,\ -3\cdot4-6\cdot3,\ 0-8)=(6,-30,-8)$, magnitude $\sqrt{36+900+64}=\sqrt{1000}$. Dividing by 5 gives $\sqrt{40}=2\sqrt{10}$, option (B).

Final Answer:
$AM=2\sqrt{10}$, option (B). \[ \boxed{2\sqrt{10}} \]