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Joint Entrance Examination
Mathematics
3D Geometry
Question 291
Question 291 of 521
Medium
NUMERICAL
If the lines
x
−
k
1
=
y
−
2
2
=
z
−
3
3
\dfrac{x-k}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}
1
x
−
k
=
2
y
−
2
=
3
z
−
3
and
x
+
1
3
=
y
+
2
2
=
z
+
3
1
\dfrac{x+1}{3}=\dfrac{y+2}{2}=\dfrac{z+3}{1}
3
x
+
1
=
2
y
+
2
=
1
z
+
3
are coplanar, then the value of
k
k
k
is: Answer: 1
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Official solution
\[\begin{vmatrix}k+1&4&6\\1&2&3\\3&2&1\end{vmatrix}=0.\] \[(k+1)(2-6)-4(1-9)+6(2-6)=0 \implies -4(k+1)+32-24=0 \implies k=1.\]
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