Exams Joint Entrance Examination Mathematics 3D Geometry Question 53 Question 53 of 521 Medium MCQ
Let P be the foot of the perpendicular from the point Q ( 10 , − 3 , − 1 ) Q(10,-3,-1) Q ( 10 , − 3 , − 1 ) on the line x − 3 7 = y − 2 − 1 = z + 1 − 2 \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2} 7 x − 3 = − 1 y − 2 = − 2 z + 1 . Then the area of the right angled triangle P Q R PQR P QR , where R R R is the point ( 3 , − 2 , 1 ) (3,-2,1) ( 3 , − 2 , 1 ) , is
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\[ \begin{aligned}
&\begin{aligned}
& \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}=\lambda \\
& \Rightarrow 7 \lambda+3,-\lambda+2,-2 \lambda-1 \\
& \text { dr's of QP } \Rightarrow \\
& 7 \lambda-7,-\lambda+5,-2 \lambda
\end{aligned} \\
\&\text { Now }\\
\& ( 7 λ − 7 ) ⋅ 7 − ( − λ + 5 ) + ( 2 λ ) ⋅ 2 = 0 54 λ − 54 = 0 ⇒ λ = 1 ∴ P = ( 10 , 1 , − 3 ) P Q → = − 4 j ^ + 2 k ^ P R → = − 7 i ^ − 3 j ^ + 4 k ^ \begin{aligned}
& (7 \lambda-7) \cdot 7-(-\lambda+5)+(2 \lambda) \cdot 2=0 \\
& 54 \lambda-54=0 \Rightarrow \lambda=1 \\
& \therefore \mathrm{P}=(10,1,-3) \\
& \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{PR}}=-7 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}
\end{aligned} ( 7 λ − 7 ) ⋅ 7 − ( − λ + 5 ) + ( 2 λ ) ⋅ 2 = 0 54 λ − 54 = 0 ⇒ λ = 1 ∴ P = ( 10 , 1 , − 3 ) PQ = − 4 j ^ + 2 k ^ PR = − 7 i ^ − 3 j ^ + 4 k ^
\end{aligned}\]
Area \[=\left| {{1 \over 2}\left| { ( i j k 0 − 4 2 − 7 − 3 4 ) \begin{pmatrix} i & j & k \\
0 & { - 4} & 2 \\
{ - 7} & { - 3} & 4 \\
\end{pmatrix} i 0 − 7 j − 4 − 3 k 2 4 } \right|} \right| = 3\sqrt {30} \] Answer the question to unlock