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 the vectors forming angle \(A\).
Angle \(A\) is formed by the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
\[ \overrightarrow{AB}=B-A=(3-1,1-(-2),-3-3) \] \[ \overrightarrow{AB}=(2,3,-6) \] Also, \[ \overrightarrow{AC}=C-A=(-3-1,1-(-2),3-3) \] \[ \overrightarrow{AC}=(-4,3,0) \]

Step 2: Use the formula for cosine of angle between two vectors.
\[ \cos A=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}} {|\overrightarrow{AB}|\,|\overrightarrow{AC}|} \] Now, \[ \overrightarrow{AB}\cdot \overrightarrow{AC} =(2)(-4)+(3)(3)+(-6)(0) \] \[ =-8+9+0=1 \]

Step 3: Find magnitudes of both vectors.
\[ |\overrightarrow{AB}|=\sqrt{2^2+3^2+(-6)^2} \] \[ =\sqrt{4+9+36}=\sqrt{49}=7 \] Also, \[ |\overrightarrow{AC}|=\sqrt{(-4)^2+3^2+0^2} \] \[ =\sqrt{16+9}=\sqrt{25}=5 \]

Step 4: Substitute the values.
\[ \cos A=\frac{1}{7\times 5} \] \[ \cos A=\frac{1}{35} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{1}{35}} \]