NCERT Solutions for Class 12 Mathemetics Chapter 11 - Three-Dimensional Geometry

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Direction–angles are 90$°$, 135$°$, 45$°$.
$\therefore$ Direction–cosines are < cos 90$°$, cos 135$°$, cos 45$°$ >

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Let each directional angle be ‘$\mathrm{\alpha }$’.
$\therefore$Direction–cosines are < cos $\mathrm{\alpha }$, cos $\mathrm{\alpha }$, cos $\mathrm{\alpha }$ >.
But l² + m² + n² = 1
$\Rightarrow$cos² $\mathrm{\alpha }$+ cos² $\mathrm{\alpha }$+ cos² $\mathrm{\alpha }$= 1 $\Rightarrow$3 cos² $\mathrm{\alpha }$= 1

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Let A (2, 3, 4); B (– 1, – 2, 1) and C (5, 8, 7) be the given points.
$\therefore$ Direction-ratios of AB are :
< – 1 – 2, – 2 – 3, 1 – 4 >
i.e.< – 3, – 5, – 3 > i.e. < 3, 5, 3 >
Direction-ratios of BC are :
< 5 – (– 1), 8 – (– 2), 7 – 1 >
i.e. < 6, 10, 6 > i.e.< 3, 5, 3 >
$\Rightarrow$AB and BC have same direction-ratios
$\Rightarrow$ AB || BC.
But B is a common point.
Hence, A, B, C are collinear.

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Let A (3, 5, – 4), B (– 1, 1, 2) and C (– 5, – 5, – 2) be the vertices of ΔABC.
Now (I) Direction–ratios of AB are
< – 1 – 3, 1 – 5, 2 – (– 4) >
i.e. < – 4, – 4, 6 >
$\therefore$ Direction-cosines of AB are :

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If A (1, – 1, 2) and B (3, 4, – 2) be the given points, then the direction–ratios of AB are :
< 3 – 1, 4 – (– 1), – 2 – 2 > i.e. < 2, 5, – 4 >.
Again if C (0, 3, 2) and D (3, 5, 6) be the given points,
then the direction ratios of CD are :
< 3 – 0, 5 – 3, 6 – 2 > i.e. < 3, 2, 4 >.
Now AB is perp. to CD if :
(2) (3) + (5) (2) + (– 4) (4) = 0
[a₁a₂ + b₁b₂ + c₁c₂ = 0]
i.e. if 6 + 10 – 16 = 0, which is true.
Hence, AB is perpendicular to CD.

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If A (4, 7, 8) and B (2, 3, 4) be the given points, then the direction-ratios of AB are :
< 2 – 4, 3 – 7, 4 – 8 > i.e. < – 2, – 4, – 4 >
i.e. < 1, 2, 2 >.
If C (– 1, – 2, 1) and D (1, 2, 5) be the given points, then
the direction–ratios of CD are :
< 1 – (– 1), 2 – (– 2), 5 – 1 > i.e. < 2, 4, 4 >
i.e. < 1, 2, 2 >.
AB is parallel to CD if :

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The given line is :

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(i) The vector equations of the line is :

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The given plane is z = 2 ....(1)
Direction-cosines of the normal to the plane are < 0, 0, 1 >.
And distance of (1) from the origin

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The given plane is x + y + z = 1 ....(1)
Direction-ratios of the normal to the plane are : < 1, 1, 1 >
$\therefore$ Direction-cosines of the normal to the plane are

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The given plane is 2x + 3y – z = 5 ....(1)
Direction–ratios of the normal to the plane are < 2, 3, – 1 >.
$\therefore$ Direction–cosines of the normal to the plane are :

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The given plane is 5y + 8 = 0 ....(1)
Direction–ratios of the normal to the plane are < 0, 5, 0 >
i.e. < 0, 1, 0 >
$\therefore$ Direction–cosines of the normal to the plane are < 0, 1, 0 > .

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The given plane is :
2x + 3y + 4z – 12 = 0 ...(1)
Direction-ratios of the normal to the plane are :
< 2, 3, 4 >
$\therefore$ The equations of the normal through (0, 0, 0) are :

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The given plane is 3y + 4z – 6 = 0 ....(1)
Direction-ratios of the normal are < 0, 3, 4 >.
$\therefore$ The equations of the normal though (0, 0, 0) are :

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The given plane is :
x + y + z = 1 ....(1)
Direction-ratios of the normal to the plane are < 1, 1, 1 >.

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The given plane is :
5y + 8 = 0 ...(1)
Direction-ratios of the normal to the plane < 0, 5, 0 >
$\therefore$ The equations of the normal through (0, 0, 0) are :

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Any plane through (1, 1, – 1) is :
a (x – 1) + b (y – 1) + c (z + 1) = 0....(1)
Since the plane passes through the points
(6, 4, – 5) and (– 4, – 2, 3),
$\therefore$ a (6 – 1) + b (4 – 1) + c (– 5 + 1) = 0
and a (– 4 – 1) + b (– 2 – 1) + c (3 + 1) = 0
$\Rightarrow$ 5a + 3b – 4c = 0 ....(2)
and – 5a – 3b + 4c = 0 ....(3)
Solving (2) and (3),

$\Rightarrow$ a, b, c can’t be found.
Since the points are collinear,
$\therefore$ an infinite number of planes can be found through the given points.

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Any plane through (1, 1, 0) is :
a (x – 1) + b (y – 1) + cz = 0 ....(1)
Since the plane passes through the points
(1, 2, 1) and (– 2, 2, – 1),
$\therefore$ a (1 – 1) + b (2 – 1) + c (1) = 0
and a (– 2 – 1) + b (2 – 1) + c (– 1) = 0
$\Rightarrow$ 0 . a + b + c = 0 ....(2)
and – 3a + b – c = 0 ....(3)
Solving (2) and (3),

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The given equation of the plane is :
2x + y – z = 5

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Clearly the plane passes through the point (0, 3, 0).
The direction-cosines of the normal to the plane are < 0, 1, 0 >
$\therefore$ The equation of the plane is :
0 . (x – 0) + 1 . (y – 3) + 0 . (z – 0) = 0
$\Rightarrow$ y – 3 = 0 $\Rightarrow$ y = 3.

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The given planes are :
3x – y + 2z – 4 = 0 ...(1)
and x + y + z – 2 = 0 ...(2)
Any plane through the intersection of (1) and (2) is :
(3x – y + 2z – 4) + k (x + y + z – 2) = 0 ...(3)
Since it passes thro’ (2, 2, 1).
$\therefore$ (6 – 2 + 2 – 4) + k (2 + 2 + 1 – 2) = 0

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The given planes are :

i.e. 2x + 2y – 3z – 7 = 0 ...(1)′
and 2x + 5y + 3z – 9 = 0 ...(2)′
Any plane through the intesection of (1)′ and (2)′ is :
(2x + 2y – 3z – 7) + k (2x + 5y + 3z – 9) = 9...(3)
Since it passes thro’ (2, 1, 3),
$\therefore$ (4 + 2 – 9 – 7) + k (4 + 5 + 9 – 9) = 0

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Any plane through the line of intersection of the planes :
x + y + z – 1 = 0 and 2x + 3y + 4z – 5 = 0 is :
(x + y + z – 1) + k (2x + 3y + 4z – 5) = 0
i.e.(1 + 2k) x + (1 + 3k) y + (1 + 4k) z
– (1 + 5k) = 0 ...(1)
Since it is perpendicular to the plane :
x – y + z = 0 ...(2)
$\therefore$ their normals are perpendicular
$\Rightarrow$ (1 + 2k) (1) + (1 + 3k) (– 1) + (1 + 4k) (1) = 0
$\Rightarrow$ 1 + 2k – 1 – 3k + 1 + 4k = 0

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The given planes are :
7x + 5y + 6z + 30 = 0 ...(1)
and 3x – y – 10z + 4 = 0 ...(2)

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The given planes are :
2x + y + 3z – 2 = 0 ....(1)
and x – 2y + 5 = 0 ....(2)
Since (2) (1) + (1) (– 2) + (3) (0) = 0,
$\therefore$ planes are parallel.

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The given planes are :
2x – 2y + 4z + 5 = 0 ....(1)
and 3x – 3y + 6z – 1 = 0 . ...(2)

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The given planes are :
2x – y + 3z – 1 = 0 ....(1)
and 2x – y + 3z + 3 = 0 ....(2)

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The given planes are :
4x + 8y + z – 8 = 0 ....(1)
and y + z – 4 = 0 ....(2)

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The given plane is :
3x – 4y + 12z – 3 = 0 ....(1)
$\therefore$ Distance, of (0, 0, 0) from plane (1)

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The given plane is :
2x – y + 2z + 3 = 0 ....(1)
$\therefore$ Distance, (3, – 2, 1) from plane (1)

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The given plane is :
x + 2y – 2z – 9 = 0 ....(1)
$\therefore$ Distance of (2, 3, – 5) from plane (1)

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The given plane is 2x – 3y + 6z – 2 = 0 ....(1)
$\therefore$ Distance of (– 6, 0, 0) from place (1)

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parallel