Show that the four points A, B, C, D with position vectors a , b … - Vidyadip

Question

Show that the four points A, B, C, D with position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},\ \vec{\text{d}}$ respectively such that $3\vec{\text{a}}-2\vec{\text{b}}+5\vec{\text{c}}-6\vec{\text{d}}=0$, are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

✓

Answer

Let AC and BD intersects at a point P.

We have,

$3\vec{\text{a}}-2\vec{\text{b}}+5\vec{\text{c}}-6\vec{\text{d}}=0$

$\Rightarrow3\vec{\text{a}}+5\vec{\text{c}}=2\vec{\text{b}}+6\vec{\text{d}}$

Since sum of co-efficients on both sides of the above equation is 8.

so we divide the equation on both sides by 8.

$\Rightarrow\frac{3\vec{\text{a}}+5\vec{\text{c}}}8=\frac{2\vec{\text{b}}+6\vec{\text{d}}}8$

$\Rightarrow\frac{3\vec{\text{a}}+5\vec{\text{c}}}{3+5}=\frac{2\vec{\text{b}}+6\vec{\text{d}}}{2+6}$

Therefore, P divides AC in the ratio of 3 : 5 and P divides BD in the ratio of 2 : 6.

Therefore, position vector of the point of intersection of AC and BD will be

$\Rightarrow\frac{3\vec{\text{a}}+5\vec{\text{c}}}8=\frac{2\vec{\text{b}}+6\vec{\text{d}}}8$

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