IIT JEE 1983 Maths- 'Adapted' - FIB to Single-Digit Integer Q5

Geometry Level 4

The unit vector perpendicular to the plane determined by P ( 1 , − 1 , 2 ) P(1,-1,2) , Q ( 2 , 0 , 1 ) Q(2,0,1) , R ( 0 , 2 , 1 ) R(0, 2,1) is 1 6 ( a i ^ + b j ^ + c k ^ ) \dfrac{1}{\sqrt6} (a\hat i+b\hat j+c\hat k) . Find ∣ a + b + c ∣ \left| a + b + c \right| .

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The answer is 4.

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

P e r p e n d i c u l a r v e c t o r = ∣ P Q → × Q R → ∣ . P Q = Q − P = ( 1 , 1 , − 1 ) , Q R = R − Q = ( − 2 , 2 , 0 ) . P e r p e n d i c u l a r v e c t o r = ∣ i j k 1 1 − 1 − 2 2 0 ∣ = ( 2 i , 2 j , 4 k ) ∴ u n i t p e r p e n d i c u l a r v e c t o r = ( 2 i , 2 j , 4 k ) 4 + 4 + 16 = ( 2 i , 2 j , 4 k ) 2 6 = ( i , j , 2 k ) 6 = ( a i , b j , c k ) 6 ∴ a + b + c ∣ = ∣ 1 + 1 + 2 ∣ = 4. Perpendicular~ vector = |\overrightarrow{PQ}\times\overrightarrow{QR}|.\\ PQ=Q - P=(1,1,-1),~~~~QR=R - Q=(-2,2,0).\\ Perpendicular~ vector= \begin{vmatrix}i&j&k\\ 1&1&-1\\ -2&2&0\\ \end{vmatrix}= (2i,2j,4k)\\ \therefore~unit~perpendicular~ vector=\dfrac{(2i,2j,4k)}{\sqrt{4+4+16}}\\ =\dfrac{(2i,2j,4k)}{2\sqrt6}=\dfrac{(i,j,2k)}{\sqrt6}\\ =\dfrac{(ai,bj,ck)}{\sqrt6}\\ \therefore~~a+b+c|=|1+1+2|=\Large \color{#D61F06}{4}.

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