If \( \hat{\imath}+\hat{\jmath}+\hat{k}, 2 \hat{\imath}+5 \hat{\jmath}

1.	If \( \hat{\imath}+\hat{\jmath}+\hat{k}, 2 \hat{\imath}+5 \hat{\jmath}, 3 \hat{\imath}+2 \hat{\jmath}-3 \hat{k} \) and \( \hat{\imath}-6 \hat{\jmath}-\hat{k} \) respectively are the position vectors of points \( A, B, C \) and \( D \), then find the angle between the straight lines \( A B \) and \( C D \). Find whether \( \overrightarrow{A B} \) and \( \overrightarrow{C D} \) are collinear or not.
Answer» Position vectors of \(\vec A\,\&\,\vec B\) are \(\hat i+\hat j + \hat k\) and \(2\hat i+5\hat j\) respectively. \(\therefore\) Direction vector of line AB is \(\vec a=(2\hat i+5\hat j)-(\hat i+\hat j+\hat k)\) ⇒ \(\vec a=\hat i+4\hat j-\hat k\) Position vectors of \(\vec C\,\&\,\vec D\) are \(3\hat i+2\hat j - 3\hat k\) and \(\hat i-6\hat j - \hat k\) respectively. \(\therefore\) Direction vector of line CD is \(\vec b=(\hat i-6\hat j-\hat k)-(3\hat i+2\hat j-3\hat k)\) ⇒ \(\vec b=-2\hat i-8\hat j+2\hat k\) Let angle between line AB and CD is θ. \(\therefore\) cos θ = \(\frac{\vec a.\vec b}{\|\vec a\|\|\vec b\|}\) ⇒ cos θ = \(\left\|\frac{(\hat i+4\hat j-\hat k).(-2\hat i-8\hat j+2\hat k)}{\sqrt{1^2+4^2+(-1)^2}\sqrt{(-2)^2+(-8)^2+2^2}}\right\|\) \(=\left\|\frac{-2-32-2}{\sqrt{18}\sqrt{72}}\right\|\) \(=\left\|\frac{-36}{\sqrt{18}\sqrt{18}.\sqrt4}\right\|\) \(=\frac{36}{18\times2}\) = 1 = cos 0° \(\therefore\) θ = 0° Hence, lines/vectors \(\vec {AB}\) and \(\vec {CD}\) are collinear.

If \( \hat{\imath}+\hat{\jmath}+\hat{k}, 2 \hat{\imath}+5 \hat{\jmath}, 3 \hat{\imath}+2 \hat{\jmath}-3 \hat{k} \) and \( \hat{\imath}-6 \hat{\jmath}-\hat{k} \) respectively are the position vectors of points \( A, B, C \) and \( D \), then find the angle between the straight lines \( A B \) and \( C D \). Find whether \( \overrightarrow{A B} \) and \( \overrightarrow{C D} \) are collinear or not.

Answer»

Position vectors of \(\vec A\,\&\,\vec B\) are \(\hat i+\hat j + \hat k\) and \(2\hat i+5\hat j\) respectively.

\(\therefore\) Direction vector of line AB is \(\vec a=(2\hat i+5\hat j)-(\hat i+\hat j+\hat k)\)

⇒ \(\vec a=\hat i+4\hat j-\hat k\)

Position vectors of \(\vec C\,\&\,\vec D\) are \(3\hat i+2\hat j - 3\hat k\) and \(\hat i-6\hat j - \hat k\) respectively.

\(\therefore\) Direction vector of line CD is

\(\vec b=(\hat i-6\hat j-\hat k)-(3\hat i+2\hat j-3\hat k)\)

⇒ \(\vec b=-2\hat i-8\hat j+2\hat k\)

Let angle between line AB and CD is θ.

\(\therefore\) cos θ = \(\frac{\vec a.\vec b}{|\vec a||\vec b|}\)

⇒ cos θ = \(\left|\frac{(\hat i+4\hat j-\hat k).(-2\hat i-8\hat j+2\hat k)}{\sqrt{1^2+4^2+(-1)^2}\sqrt{(-2)^2+(-8)^2+2^2}}\right|\)

\(=\left|\frac{-2-32-2}{\sqrt{18}\sqrt{72}}\right|\)

\(=\left|\frac{-36}{\sqrt{18}\sqrt{18}.\sqrt4}\right|\)

\(=\frac{36}{18\times2}\) = 1 = cos 0°

\(\therefore\) θ = 0°

Hence, lines/vectors \(\vec {AB}\) and \(\vec {CD}\) are collinear.