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Class 11thExemplar SolutionMathsMaths NCERT Exemplar Solutions Class 11th Chapter TwelveIntroduction to Three Dimensional Geometry

Match each item given under Column C1 to its correct answer given under Column C2:

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This is a matching answer type question as classified in NCERT Exemplar

(a)In  xy-plane,  z-coordinate  is  zero.             H e n c e ,     ( a ) ↔ ( i i i ) . (b)The  point  (2,3,4)  lies  in  first  octant.             H e n c e ,     ( b ) ↔ ( i ) . (c)Locus  of  the  points  with  x-coordinate  is  zero  is  yz-plane.             H e n c e ,     ( c ) ↔ ( i i ) . (d)A  line  is  parallel  to  x-axis  if  and  only  if  all  the  points  on  the  line        have  equal  y  and  z-coordinates.             H e n c e ,     ( d ) ↔ ( v i ) . (e)x=0,  y=0  represents  z-axis.             H e n c e ,     ( e ) ↔ ( i v ) . (f)z=c  represent  a  plane  parallel  to  xy-plane.             H e n c e ,     ( f ) ↔ ( v ) . (g)The  planes  x=a,  y=b  represent  the  line  parallel  to  z-axis.             H e n c e ,     ( g ) ↔ ( v i i i ) . (h)Coordinates  of  a  point  are  the  distances  from  the  origin  to  the  feet  of  perpendicular  from         the  point  on  the  respective  axes,             H e n c e ,     ( h ) ↔ ( v i i ) . ( i ) A     b a l l     i s     s o l i d     r e g i o n     i n     t h e     s p a c e     e n c l o s e d     b y     a     s p h e r e .             H e n c e ,     ( i ) ↔ ( x ) . ( j ) T h e     r e g i o n     i n     t h e     p l a n e     e n c l o s e d     b y     a     c i r c l e     i s     k n o w n     a s     a     d i s c .             H e n c e ,     ( i ) ↔ ( i x ) .

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The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,-4,-5) and R(2,-3,1) and the plane 2x + y + z = 7, is equal to………..

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Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7

For real numbers α and β ≠ 0, if the point of intersection of the straight lines (x-α)/1 = (y-1)/2 = (z-1)/3 and (x-4)/β = (y-6)/3 = (z-7)/3 lies on the plane x+2y-z = 8, then α - β is equal to:

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Any point on line (1)
x=α+k
y=1+2k
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⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
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⇒α+1+6−4=8 . (i)
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Let f : [0,∞) → be a function defined by f(x) = { max{sint: 0≤ t ≤x}, 0≤x≤π, 2 + cos x, x > π
Then which of the following is true?
<!-- [if !supportLineBreakNewLine]-->
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f (x)= {sinx, 0≤x<π/2; 1, π/2≤x≤π 2+cosx, x>π}
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f' (π? ) = 0
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The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3x + y – 2z = 5 and

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Let direction ratio of the normal to the required plane are l, m, n

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The distance of the point (1, 1, 9) form the point of intersection of the line x − 3 1 = y − 4 2 = z − 5 2 <!-- [if gte mso 9]><xml> </xml><![endif]--> and the plane  x + y + z = 17 is:

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Any point on line  x − 3 1 = y − 4 2 = z − 5 2 = r is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve 2025

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve 2025

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