Question

If the position vectors of points \(A\), \(B\), \(C\) and \(D\) are respectively \(\hat i+\hat j+\hat k\), \(2\hat i+5\hat j\), \(3\hat i+2\hat j-3\hat k\) and \(\hat i-6\hat j-\hat k\), then find the angle between the lines \(AB\) and \(CD\). Prove that \(AB\) and \(CD\) are collinear.

Hint:

Find vectors AB and CD by subtraction, then check if one is a scalar multiple of the other.

Answer 1

Step 1: Understanding the Concept:
The vector along the line joining two points is found by subtracting the position vector of the starting point from that of the end point.
The angle between two lines is the angle between their direction vectors, found using the dot product formula \(\cos\theta=\dfrac{\vec{AB}\cdot\vec{CD}}{|\vec{AB}||\vec{CD}|}\).

Step 2: Find the Vectors AB and CD:
Position vectors are \(A=\hat i+\hat j+\hat k\), \(B=2\hat i+5\hat j\), \(C=3\hat i+2\hat j-3\hat k\), \(D=\hat i-6\hat j-\hat k\).
Subtract to get the direction of each line.
\[ \vec{AB}=B-A=\hat i+4\hat j-\hat k, \qquad \vec{CD}=D-C=-2\hat i-8\hat j+2\hat k \]

Step 3: Compute the Angle:
Find the magnitudes and the dot product of the two vectors.
\[ |\vec{AB}|=\sqrt{1+16+1}=3\sqrt2, \qquad |\vec{CD}|=\sqrt{4+64+4}=6\sqrt2 \]
\[ \vec{AB}\cdot\vec{CD}=(1)(-2)+(4)(-8)+(-1)(2)=-36 \]
\[ \cos\theta=\frac{-36}{3\sqrt2\times6\sqrt2}=\frac{-36}{36}=-1 \implies \theta=180^\circ \]

Step 4: Prove AB and CD are Collinear:
Compare the two vectors found in Step 2.
\[ \vec{CD}=-2\hat i-8\hat j+2\hat k=-2(\hat i+4\hat j-\hat k)=-2\,\vec{AB} \]
Since \(\vec{CD}\) is a scalar multiple of \(\vec{AB}\), the two vectors are parallel, which means lines AB and CD are collinear.
The scalar \(-2\) is negative, so the vectors point in opposite senses, which is exactly why the angle between them came out as \(180^\circ\) in Step 3.

Final Answer:
The angle between AB and CD is 180 degrees, and CD = -2 AB proves the lines are collinear.
\[ \boxed{\theta=180^\circ, \quad \vec{CD}=-2\vec{AB}} \]