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

Let AC and BD intersects at a point P. We have, 3 a -2 b +5 c -6 d =0 3 a +5 c =2 b +6 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. 3 a +5 c 8= 2 b +6 d 8 3 a +5 c 3+5 = 2 b +6 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 3 a +5 c 8= 2 b +6 d 8

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

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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.

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