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Exam: Maharashtra HSC Class 12
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Q901
#901
Mathematics → Line and Planes → Vector and Cartesian Equations of a Line
·
Hard
Add
Show that lines
x
+
1
−
10
=
y
+
3
−
1
=
z
−
4
1
\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}
−
10
x
+
1
=
−
1
y
+
3
=
1
z
−
4
and
x
+
10
−
1
=
y
+
1
−
3
=
z
−
1
4
\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}
−
1
x
+
10
=
−
3
y
+
1
=
4
z
−
1
intersect each other. Find the co-ordinates of their point of intersection.
Show model answer
[QEx 6.1 Q.9 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q902
#902
Mathematics → Line and Planes → Vector and Cartesian Equations of a Line
·
Moderate
Add
A line passes through
(
3
,
−
1
,
2
)
(3,-1,2)
(
3
,
−
1
,
2
)
and is perpendicular to lines
r
ˉ
=
(
i
^
+
j
^
−
k
^
)
+
λ
(
2
i
^
−
2
j
^
+
k
^
)
\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+\hat{k})
r
ˉ
=
(
i
^
+
j
^
−
k
^
)
+
λ
(
2
i
^
−
2
j
^
+
k
^
)
and
r
ˉ
=
(
2
i
^
+
j
^
−
3
k
^
)
+
μ
(
i
^
−
2
j
^
+
2
k
^
)
\bar{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})
r
ˉ
=
(
2
i
^
+
j
^
−
3
k
^
)
+
μ
(
i
^
−
2
j
^
+
2
k
^
)
. Find its equation.
Show model answer
[QEx 6.1 Q.10 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q903
#903
Mathematics → Line and Planes → Vector and Cartesian Equations of a Line
·
Moderate
Add
Show that the line
x
−
2
1
=
y
−
4
2
=
z
+
4
−
2
\frac{x-2}{1}=\frac{y-4}{2}=\frac{z+4}{-2}
1
x
−
2
=
2
y
−
4
=
−
2
z
+
4
passes through the origin.
Show model answer
[QEx 6.1 Q.11 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q904
#904
Mathematics → Line and Planes → Distance of a Point from a Line
·
Moderate
Add
Find the length of the perpendicular drawn from the point
P
(
3
,
2
,
1
)
P(3,2,1)
P
(
3
,
2
,
1
)
to the line
r
ˉ
=
(
7
i
^
+
7
j
^
+
6
k
^
)
+
λ
(
−
2
i
^
+
2
j
^
+
3
k
^
)
\bar{r}=\left(7\hat{i}+7\hat{j}+6\hat{k}\right)+\lambda\left(-2\hat{i}+2\hat{j}+3\hat{k}\right)
r
ˉ
=
(
7
i
^
+
7
j
^
+
6
k
^
)
+
λ
(
−
2
i
^
+
2
j
^
+
3
k
^
)
.
Show model answer
[Q6.2 SolvedEx.12 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q905
#905
Mathematics → Line and Planes → Distance of a Point from a Line
·
Moderate
Add
Find the distance of the point
P
(
0
,
2
,
3
)
P(0,2,3)
P
(
0
,
2
,
3
)
from the line
x
+
3
5
=
y
−
1
2
=
z
+
4
3
\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}
5
x
+
3
=
2
y
−
1
=
3
z
+
4
.
Show model answer
[Q6.2 SolvedEx.13 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q906
#906
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Find the shortest distance between lines
r
ˉ
=
(
2
i
^
−
j
^
)
+
λ
(
2
i
^
+
j
^
−
3
k
^
)
\bar{r}=(2\hat{i}-\hat{j})+\lambda(2\hat{i}+\hat{j}-3\hat{k})
r
ˉ
=
(
2
i
^
−
j
^
)
+
λ
(
2
i
^
+
j
^
−
3
k
^
)
and
r
ˉ
=
(
i
^
−
j
^
+
2
k
^
)
+
μ
(
2
i
^
+
j
^
−
5
k
^
)
\bar{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}+\hat{j}-5\hat{k})
r
ˉ
=
(
i
^
−
j
^
+
2
k
^
)
+
μ
(
2
i
^
+
j
^
−
5
k
^
)
.
Show model answer
[Q6.3 SolvedEx.14 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q907
#907
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Find the shortest distance between lines
x
−
1
2
=
y
−
2
3
=
z
−
3
4
\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}
2
x
−
1
=
3
y
−
2
=
4
z
−
3
and
x
−
2
3
=
y
−
4
4
=
z
−
5
5
\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}
3
x
−
2
=
4
y
−
4
=
5
z
−
5
.
Show model answer
[Q6.3 SolvedEx.15 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q908
#908
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Show that lines
r
ˉ
=
(
i
^
+
j
^
−
k
^
)
+
λ
(
2
i
^
−
2
j
^
+
k
^
)
\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+\hat{k})
r
ˉ
=
(
i
^
+
j
^
−
k
^
)
+
λ
(
2
i
^
−
2
j
^
+
k
^
)
and
r
ˉ
=
(
4
i
^
−
3
j
^
+
2
k
^
)
+
μ
(
i
^
−
2
j
^
+
2
k
^
)
\bar{r}=(4\hat{i}-3\hat{j}+2\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})
r
ˉ
=
(
4
i
^
−
3
j
^
+
2
k
^
)
+
μ
(
i
^
−
2
j
^
+
2
k
^
)
intersect each other.
Show model answer
[Q6.3 SolvedEx.16 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q909
#909
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Find the distance between parallel lines
r
ˉ
=
(
2
i
^
−
j
^
+
k
^
)
+
λ
(
2
i
^
+
j
^
−
2
k
^
)
\bar{r}=(2\hat{i}-\hat{j}+\hat{k})+\lambda(2\hat{i}+\hat{j}-2\hat{k})
r
ˉ
=
(
2
i
^
−
j
^
+
k
^
)
+
λ
(
2
i
^
+
j
^
−
2
k
^
)
and
r
ˉ
=
(
i
^
−
j
^
+
2
k
^
)
+
μ
(
2
i
^
+
j
^
−
2
k
^
)
\bar{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}+\hat{j}-2\hat{k})
r
ˉ
=
(
i
^
−
j
^
+
2
k
^
)
+
μ
(
2
i
^
+
j
^
−
2
k
^
)
.
Show model answer
[Q6.3 SolvedEx.17 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q910
#910
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Find the distance between parallel lines
x
2
=
y
−
1
=
z
2
\dfrac{x}{2}=\dfrac{y}{-1}=\dfrac{z}{2}
2
x
=
−
1
y
=
2
z
and
x
−
1
2
=
y
−
1
−
1
=
z
−
1
2
\dfrac{x-1}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}
2
x
−
1
=
−
1
y
−
1
=
2
z
−
1
.
Show model answer
[Q6.3 SolvedEx.18 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q911
#911
Mathematics → Line and Planes → Distance of a Point from a Line
·
Moderate
Add
Find the length of the perpendicular from
(
2
,
−
3
,
1
)
(2,-3,1)
(
2
,
−
3
,
1
)
to the line
x
+
1
2
=
y
−
3
3
=
z
+
1
−
1
\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z+1}{-1}
2
x
+
1
=
3
y
−
3
=
−
1
z
+
1
.
Show model answer
[QEx 6.2 Q.1 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q912
#912
Mathematics → Line and Planes → Distance of a Point from a Line
·
Moderate
Add
Find the co-ordinates of the foot of the perpendicular drawn from the point
2
i
^
−
j
^
+
5
k
^
2\hat{i}-\hat{j}+5\hat{k}
2
i
^
−
j
^
+
5
k
^
to the line
r
ˉ
=
(
11
i
^
−
2
j
^
−
8
k
^
)
+
λ
(
10
i
^
−
4
j
^
−
11
k
^
)
\bar{r}=\left(11\hat{i}-2\hat{j}-8\hat{k}\right)+\lambda\left(10\hat{i}-4\hat{j}-11\hat{k}\right)
r
ˉ
=
(
11
i
^
−
2
j
^
−
8
k
^
)
+
λ
(
10
i
^
−
4
j
^
−
11
k
^
)
. Also find the length of the perpendicular.
Show model answer
[QEx 6.2 Q.2 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q913
#913
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Find the shortest distance between the lines
r
ˉ
=
(
4
i
^
−
j
^
)
+
λ
(
i
^
+
2
j
^
−
3
k
^
)
\bar{r}=\left(4\hat{i}-\hat{j}\right)+\lambda\left(\hat{i}+2\hat{j}-3\hat{k}\right)
r
ˉ
=
(
4
i
^
−
j
^
)
+
λ
(
i
^
+
2
j
^
−
3
k
^
)
and
r
ˉ
=
(
i
^
−
j
^
+
2
k
^
)
+
μ
(
i
^
+
4
j
^
−
5
k
^
)
\bar{r}=\left(\hat{i}-\hat{j}+2\hat{k}\right)+\mu\left(\hat{i}+4\hat{j}-5\hat{k}\right)
r
ˉ
=
(
i
^
−
j
^
+
2
k
^
)
+
μ
(
i
^
+
4
j
^
−
5
k
^
)
.
Show model answer
[QEx 6.2 Q.3 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q914
#914
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
Find the shortest distance between the lines
x
+
1
7
=
y
+
1
−
6
=
z
+
1
1
\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}
7
x
+
1
=
−
6
y
+
1
=
1
z
+
1
and
x
−
3
1
=
y
−
5
−
2
=
z
−
7
1
\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}
1
x
−
3
=
−
2
y
−
5
=
1
z
−
7
.
Show model answer
[QEx 6.2 Q.4 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q915
#915
Mathematics → Line and Planes → Distance of a Point from a Line
·
Moderate
Add
Find the perpendicular distance of the point
(
1
,
0
,
0
)
(1,0,0)
(
1
,
0
,
0
)
from the line
x
−
1
2
=
y
+
1
−
3
=
z
+
10
8
\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+10}{8}
2
x
−
1
=
−
3
y
+
1
=
8
z
+
10
. Also find the co-ordinates of the foot of the perpendicular.
Show model answer
[QEx 6.2 Q.5 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q916
#916
Mathematics → Line and Planes → Distance of a Point from a Line
·
Moderate
Add
A
(
1
,
0
,
4
)
A(1,0,4)
A
(
1
,
0
,
4
)
,
B
(
0
,
−
11
,
13
)
B(0,-11,13)
B
(
0
,
−
11
,
13
)
,
C
(
2
,
−
3
,
1
)
C(2,-3,1)
C
(
2
,
−
3
,
1
)
are three points and
D
D
D
is the foot of the perpendicular from
A
A
A
to
B
C
BC
B
C
. Find the co-ordinates of
D
D
D
.
Show model answer
[QEx 6.2 Q.6 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Set · 2 questions
By computing the shortest distance, determine whether following lines intersect each other.
Q917
#917
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
r
ˉ
=
(
i
^
−
j
^
)
+
λ
(
2
i
^
+
k
^
)
\bar{r}=\left(\hat{i}-\hat{j}\right)+\lambda\left(2\hat{i}+\hat{k}\right)
r
ˉ
=
(
i
^
−
j
^
)
+
λ
(
2
i
^
+
k
^
)
and
r
ˉ
=
(
2
i
^
−
j
^
)
+
μ
(
i
^
+
j
^
−
k
^
)
\bar{r}=\left(2\hat{i}-\hat{j}\right)+\mu\left(\hat{i}+\hat{j}-\hat{k}\right)
r
ˉ
=
(
2
i
^
−
j
^
)
+
μ
(
i
^
+
j
^
−
k
^
)
Show model answer
[QEx 6.2 Q.7 i) · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q918
#918
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Moderate
Add
x
−
5
4
=
y
−
7
−
5
=
z
+
3
−
5
\dfrac{x-5}{4}=\dfrac{y-7}{-5}=\dfrac{z+3}{-5}
4
x
−
5
=
−
5
y
−
7
=
−
5
z
+
3
and
x
−
8
7
=
y
−
7
1
=
z
−
5
3
\dfrac{x-8}{7}=\dfrac{y-7}{1}=\dfrac{z-5}{3}
7
x
−
8
=
1
y
−
7
=
3
z
−
5
Show model answer
[QEx 6.2 Q.7 ii) · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q919
#919
Mathematics → Line and Planes → Skew Lines and Shortest Distance
·
Hard
Add
If lines
x
−
1
2
=
y
+
1
3
=
z
−
1
4
\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}
2
x
−
1
=
3
y
+
1
=
4
z
−
1
and
x
−
3
1
=
y
−
k
2
=
z
1
\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}
1
x
−
3
=
2
y
−
k
=
1
z
intersect each other then find
k
k
k
.
Show model answer
[QEx 6.2 Q.8 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q920
#920
Mathematics → Line and Planes → Equations of a Plane
·
Easy
Add
Find the vector equation of the plane passing through the point having position vector
2
i
^
+
3
j
^
+
4
k
^
2\hat{i}+3\hat{j}+4\hat{k}
2
i
^
+
3
j
^
+
4
k
^
and perpendicular to the vector
2
i
^
+
j
^
−
2
k
^
2\hat{i}+\hat{j}-2\hat{k}
2
i
^
+
j
^
−
2
k
^
.
Show model answer
[Q6.4 SolvedEx.1 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q921
#921
Mathematics → Line and Planes → Equations of a Plane
·
Easy
Add
Find the Cartesian equation of the plane passing through
A
(
1
,
2
,
3
)
A(1, 2, 3)
A
(
1
,
2
,
3
)
and the direction ratios of whose normal are
3
,
2
,
5
3, 2, 5
3
,
2
,
5
.
Show model answer
[Q6.4 SolvedEx.2 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q922
#922
Mathematics → Line and Planes → Equations of a Plane
·
Moderate
Add
The foot of the perpendicular drawn from the origin to a plane is
M
(
2
,
1
,
−
2
)
M(2, 1, -2)
M
(
2
,
1
,
−
2
)
. Find the vector equation of the plane.
Show model answer
[Q6.4 SolvedEx.3 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q923
#923
Mathematics → Line and Planes → Equations of a Plane
·
Moderate
Add
Find the vector equation of the plane passing through the point
A
(
−
1
,
2
,
−
5
)
A(-1, 2, -5)
A
(
−
1
,
2
,
−
5
)
and parallel to vectors
4
i
^
−
j
^
+
3
k
^
4\hat{i}-\hat{j}+3\hat{k}
4
i
^
−
j
^
+
3
k
^
and
i
^
+
j
^
−
k
^
\hat{i}+\hat{j}-\hat{k}
i
^
+
j
^
−
k
^
.
Show model answer
[Q6.4 SolvedEx.4 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q924
#924
Mathematics → Line and Planes → Equations of a Plane
·
Moderate
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Find the Cartesian equation of the plane
r
ˉ
=
(
i
^
−
j
^
)
+
λ
(
i
^
+
j
^
+
k
^
)
+
μ
(
i
^
−
2
j
^
+
3
k
^
)
\bar{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})
r
ˉ
=
(
i
^
−
j
^
)
+
λ
(
i
^
+
j
^
+
k
^
)
+
μ
(
i
^
−
2
j
^
+
3
k
^
)
.
Show model answer
[Q6.4 SolvedEx.5 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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Q925
#925
Mathematics → Line and Planes → Equations of a Plane
·
Hard
Add
Find the vector equation of the plane passing through points
A
(
1
,
1
,
2
)
A(1, 1, 2)
A
(
1
,
1
,
2
)
,
B
(
0
,
2
,
3
)
B(0, 2, 3)
B
(
0
,
2
,
3
)
and
C
(
4
,
5
,
6
)
C(4, 5, 6)
C
(
4
,
5
,
6
)
.
Show model answer
[Q6.4 SolvedEx.6 · Maharashtra State Board (Class 12) — Line and Planes (Balbharati textbook)]
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