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Vector Algebra

Question
CBSEENMA12033964

ABC is a triangle and P any point in BC. If PQ with rightwards arrow on top is the sum of AP with rightwards arrow on top comma space PB with rightwards arrow on top comma space PC with rightwards arrow on top comma show that ABQC is a || gm and Q. therefore a fixed point.

Solution

Here PQ with rightwards arrow on top space equals space AP with rightwards arrow on top space plus space PB with rightwards arrow on top space plus space PC with rightwards arrow on top                                  (given)
therefore space space space PQ with rightwards arrow on top space minus space PC with rightwards arrow on top space equals space AP with rightwards arrow on top space plus space PB with rightwards arrow on top
therefore space space PQ with rightwards arrow on top plus space CP with rightwards arrow on top space equals space AP with rightwards arrow on top space plus space PB with rightwards arrow on top
therefore space space space space space CQ with rightwards arrow on top space equals space AB with rightwards arrow on top space space space space space space space space space space space space rightwards double arrow space space space space CQ space equals space AB space and space CQ space vertical line vertical line thin space AB
∴ ABQC is a parallelogram
But A. B, C are given to be fixed points and ABQC is a || gm
∴ Q is a fixed point.