Question

Suppose \((l_1,m_1,n_1)\) and \((l_2,m_2,n_2)\) are the direction cosines of two lines and \(\theta\) is the angle between them such that \[ \cos\theta=\pm(l_1l_2+m_1m_2+n_1n_2) \] Let \(A=(1,-2,3)\), \(B=(3,1,-3)\) and \(C=(-3,1,3)\) be the vertices of a triangle \(ABC\). Then \(\cos A=\)

• A. \(-\dfrac{1}{35}\)
• B. \(\dfrac{1}{7}\)
• C. \(-\dfrac{1}{7}\)
• D. \(\dfrac{1}{35}\)

Hint:

For finding the angle at a vertex of a triangle in 3D geometry, always form two vectors starting from that vertex and then use the dot product formula.

Answer 1

The Correct Option is D

Step 1: Find vectors AB and AC.
\(A=(1,-2,3)\), \(B=(3,1,-3)\), \(C=(-3,1,3)\). \(\overrightarrow{AB}=(2,3,-6)\), \(\overrightarrow{AC}=(-4,3,0)\).

Step 2: Use dot product to find cos A.
\(\overrightarrow{AB}\cdot\overrightarrow{AC}=2(-4)+3(3)+(-6)(0)=-8+9=1\). \(|\overrightarrow{AB}|=7\), \(|\overrightarrow{AC}|=5\). \[\cos A=\frac{1}{35}=\boxed{\dfrac{1}{35}}\]