The position vector of four points P - Q - R - S are respectively 2 -k- 5 -3 -3-4 k-2 -3-k- and 2 -k-Prove that PQ and RS are parallel and PQ -3-2 RS

P⃗=2î+4k̂ Q⃗=5î+3√(3)î+4k̂ R⃗=2√(3)î+k̂ S⃗=2î+k̂P⃗Q⃗=(5 2)î+3√(3)ĵ+(1 4)k̂ R⃗S⃗2î+2√(3)ĵ+(1 1)k̂ =2î+2√(3)=2(i+√(3)j) Hence the direction of both PQ amp; RS ...

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The position vector of four points P,Q,R,S are respectively $2^i +^k, 5^i + 3 \sqrt{3}^j + 4^k, - 2 \sqrt{3}^j +^k a n d 2^i +^k$
Prove that PQ and RS are parallel and $P Q = \frac{3}{2} R S$

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Solution

$\to P = 2^i + 4^k \to Q = 5^i + 3 \sqrt{3}^i + 4^k \to R = 2 \sqrt{3}^i +^k \to S = 2^i +^k \to P Q = (5 - 2)^i + 3 \sqrt{3}^j + (1 - 4)^k \to R S 2^i + 2 \sqrt{3}^j + (1 - 1)^k = 2^i + 2 \sqrt{3} = 2 (i + \sqrt{3} j)$
Hence the direction of both PQ & RS is $i + \sqrt{3} j$
$\Rightarrow$ PQ is parallel to RS.
$| P Q | = \sqrt{3^{2}} + (3 \sqrt{3})^{2} = \sqrt{9} + 27 = \sqrt{36} = 6 | R S | = {\sqrt{2^{2} + (2 \sqrt{3})}}^{2} = \sqrt{16} = 4 \Rightarrow \frac{P Q}{R S} - \frac{6}{4} = \frac{3}{2}$

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The position vector of four points P,Q,R,S are respectively 2^i+^k,5^i+3√3^j+4^k,−2√3^j+^kand2^i+^k Prove that PQ and RS are parallel and PQ=32RS

→P=2^i+4^k→Q=5^i+3√3^i+4^k→R=2√3^i+^k→S=2^i+^k→PQ=(5−2)^i+3√3^j+(1−4)^k→RS2^i+2√3^j+(1−1)^k=2^i+2√3=2(i+√3j) Hence the direction of both PQ & RS is i+√3j ⇒ PQ is parallel to RS. |PQ|=√32+(3√3)2=√9+27=√36=6|RS|=√22+(2√3)2=√16=4⇒PQRS−64=32

The position vector of four points P,Q,R,S are respectively $2^i +^k, 5^i + 3 \sqrt{3}^j + 4^k, - 2 \sqrt{3}^j +^k a n d 2^i +^k$
Prove that PQ and RS are parallel and $P Q = \frac{3}{2} R S$