Discussion Forum

If a,b,c are  distinct  real numbers are P,Q, R are three points  whose position  vectors are  respectively 

$a\hat{i}+b\hat{j}+c\hat{k},b\hat{i}+c\hat{j}+a\hat{k}$ and $c\hat{i}+a\hat{j}+b\hat{k}$  , then $\angle QPR$=

Answer:

Explanation:

position vector of P= $a\hat{i}+b\hat{j}+c\hat{k}$

Position vector of Q= $b\hat{i}+c\hat{j}+a\hat{k}$

Position vector R=$c\hat{i}+a\hat{j}+b\hat{k}$

Now, PQ= PV of Q-PV of P

 =($b\hat{i}+c\hat{j}+a\hat{k}$)-($a\hat{i}+b\hat{j}+c\hat{k}$)

=$(b-a)\hat{i}+(c-b)\hat{j}+(a-c)\hat{k}$

 Similarly, PR= $(c-a)\hat{i}+(a-b)\hat{j}+(b-c)\hat{k}$

Now, PQ.PR=[$(b-a)\hat{i}+(c-b)\hat{j}+(a-c)\hat{k}$].[$(c-a)\hat{i}+(a-b)\hat{j}+(b-c)\hat{k}$]

    =(b-a)(c-a)+(c-b)(a-b)+(a-c)(b-c)

=$a^{2}+b^{2}+c^{2}-ab-bc-ca$

 |PQ|= $\sqrt{(b-a)^{2}+(c-b)^{2}+(a-c)^{2}}$

|PR|= $\sqrt{(c-a)^{2}+(a-b)^{2}+(b-c)^{2}}$

Now, $\cos \theta $=$\frac{PQ.PR}{|PQ|.|PR|}$

 = $\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca}$

 = $\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{2(a^{2}+b^{2}+c^{2}-ab-bc-ca)}$

$\cos \theta =\frac{1}{2}$

 $\therefore$  $\theta$= $\frac{\pi}{3}$

 Here, $\angle QPR$=$\frac{\pi}{3}$