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Lines in planes - class-XII

Description: lines in planes,matrix algebra
Number of Questions: 17
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Tags: matrix algebra maths inverse of a matrix and linear equations
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$x + 2y + 3z = 0$ 

$2x + 3y + 4z = 0$
$7x + 13y + 19z =0$
The system of equations have non trivial solutions

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. True

  2. False

Show answer
Correct Option: A

$2x - 3y + z = 0$ 

$x + 2y - 3z =0$
$4x - y - 2z = 0$
The system of equations have a non trivial solution

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. True

  2. False

Show answer
Correct Option: B

If the lines $L _{1}:\lambda ^{2}x-y-1=0$ $L _{2}:x-\lambda ^{2}y+1=0$ $L _{3}:x+y-\lambda ^{2}=0$ pass through the same point the value(s) of $\lambda$ equals

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $1$

  2. $\sqrt{2}$

  3. $2$

  4. $0$

Show answer
Correct Option: B
Explanation:

The given equations passes through same point.So, they are concurrent lines

$\Rightarrow \left| \begin{matrix} { \lambda  }^{ 2 } & -1 & -1 \ 1 & -{ \lambda  }^{ 2 } & 1 \ 1 & 1 & -{ \lambda  }^{ 2 } \end{matrix} \right| =0$

$\Rightarrow { \lambda  }^{ 6 }-3{ \lambda  }^{ 2 }-2=0$

By doing synthetic division we get,
$(\lambda^4-2\lambda^2+1)(\lambda^2-2)=0$
$(\lambda^2-1)^2(\lambda^2-2)=0$
$\lambda=\pm \sqrt 2 or \lambda=\pm1$
But here $\lambda=\pm1\  does\  not\  satisfies\  ,hence\  \lambda=\sqrt2$
Option B satisfies above equation

Three digits numbers $ 7x,36y$  and  $12z$ where  $x , y , z$ are integers from  $0$  to  $9 ,$  are divisible by a fixed constant $k.$  Then the determinant  $\left| \begin{array} { l l l } { x } & { 3 } & { 1 } \ { 7 } & { 6 } & { z } \ { 1 } & { y } & { 2 } \end{array} \right|$ $\ +48$ must be divisible by 

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $k$

  2. $k ^ { 2 }$

  3. $k ^ { 3 }$

  4. $k ^ { 4}$

Show answer
Correct Option: A
Explanation:

Since $7x,36y,12z$ are divisible by $k$

Let us Assume $x=ka,y=kb,z=kc$
So the det $\implies \left| \begin {array}{c c c} ka&3&1\7&6&kc\1&kb&2 \end{array} \right|$
$\implies 12ka-k^3abc+3kc+7kb-48$
$\implies From \  Question, 12ka-k^3abc+3kc+7kb-48+48$
Hence A

Numbers of ways in which 75600 can be resolved as product of two divisors which are relatively prime ?

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. 44

  2. $8$

  3. $9$

  4. $16$

Show answer
Correct Option: B
Explanation:

First we will have to find the prime factors of $75600$

Prime factorization of $75600=2\times 2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 7$
$\Rightarrow 75600=2^4\times 3^3 \times 5^2\times 7$
The number of ways in which a composite number N can be resolved as product of two divisors which are relatively prime.
$=2^{n-1}$ where n is number of different factors of N
$=2^{n-1}$
$=2^{4-1}$
$=2^3$
$=8$ ways
Hence, the answer is $8.$

Number of values of $a$ for which the lines $2x+y-1=0, ax+3y-3=0, 3x+2y-2=0$ are concurrent is

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. 0

  2. 1

  3. 2

  4. $\infty$

Show answer
Correct Option: D
Explanation:

Here coefficient matrix,  $\Delta = \begin{vmatrix} 2 & 1 & -1 \ a & 3 & -3 \ 3 & 2 & -2 \end{vmatrix}$
Using $C _2 \to C _2+C _3$
$\Delta = \begin{vmatrix} 2 & 0 & -1 \ a & 0 & 3 \ 3 & 0 & -2 \end{vmatrix}=0$
Clearly $\Delta=0$, Hence given lines are concurrent for all values of $a$

If the lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$ are concurrent then the value of m is

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $\displaystyle \frac{19}{5}$

  2. $\displaystyle 1$

  3. $\displaystyle \frac{5}{19}$

  4. none of these

Show answer
Correct Option: C
Explanation:

Given lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$
For concurrency,
$\begin{vmatrix} -1 & 1 & 5 \ 3 & 4 & 1 \ -m & 1 & 3 \end{vmatrix}=0$
$\Rightarrow -5+19m=0$
$\Rightarrow \displaystyle m =\frac{5}{19}$

Three lines $px + qy + r = 0, qx + ry + p = 0$ and $rx + py + q = 0$ are concurrent if

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $p + q + r = 0$

  2. $p^2 + q^2 + r^2 = pr + qr + pq$

  3. $p^3 + q^3 + r^3 = 3pqr$

  4. none of these

Show answer
Correct Option: A,B,C
Explanation:

These lines are concurrent when area formed from these lines is zero
$\begin{vmatrix} p & q & r \ q & r & p \ r & p & q \end{vmatrix}=0$
Applying ${ R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }+{ R } _{ 3 }$
$\Rightarrow \begin{vmatrix} p+q+r & p+q+r & p+q+r \ q & r & p \ r & p & q \end{vmatrix}=0\ \Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 1 & 1 \ q & r & p \ r & p & q \end{vmatrix}=0$
Again applying ${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 },{ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$
$\Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 0 & 0 \ q & r-q & p-q \ r & p-r & q-r \end{vmatrix}\ \Rightarrow \left( p+q+r \right) \left( \left( r-q \right) \left( q-r \right) -\left( p-r \right) \left( p-q \right)  \right) =0$<br>$\Rightarrow p^3+q^3+r^3-3pqr =0 $
Hence, options A, B and C are correct.

If the lines $2\mathrm{x}-\mathrm{a}\mathrm{y}+1 =0,\ 3\mathrm{x}-\mathrm{b}\mathrm{y}+1 =0,\ 4\mathrm{x}-\mathrm{c}\mathrm{y}+1 =0$ are concurrent then $\mathrm{a}$, b,c are in ?

.

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. G.P.

  2. A.P.

  3. H.P.

  4. A.G.P.

Show answer
Correct Option: B
Explanation:

$2x-ay+1 =0,\ 3x-by+1 =0,\ 4x-cy+1 =0$


Given lines are concurrent

$\Rightarrow \begin{vmatrix} 2 & -a & 1 \ 3 & -b & 1 \ 4 & -c & 1 \end{vmatrix}=0$

$\Rightarrow  2b-a-c=0$

$\Rightarrow 2b=a+c$

Hence, a,b,c are in A.P.

The points $(k, 2-2k)$, $(-k+1, 2k)$ and $(-4-k, 6-2k)$ are collinear for

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. all values of k

  2. $k=-1$

  3. $k=1/2$

  4. no value of k.

Show answer
Correct Option: B,C
Explanation:

The given points are collinear if 


$\displaystyle \begin{vmatrix} k  & 2-2k  & 1 \ -k+1  & 2k  & 1 \ -4-k & 6-2k  & 1  \end{vmatrix}=0$

$\Rightarrow  
\begin{vmatrix} k  & 2-2k  & 1 \ -2k+1  & 4k-2  & 0 \ -4-2k & 4  & 0  \end{vmatrix}=0$       $ \left [ R _{2}\rightarrow R _{2}-R _{1}, R _{3}\rightarrow R _{3}-R _{1} \right ]$


$\Rightarrow 4(-2k+1)-(-4-2k)(4k-2)=0$

$\Rightarrow (1-2k)(4-8-4k)=0$

$\Rightarrow (1-2k)(k+1)=0$

$\Rightarrow k=-1 \ or \ k=1/2$

The straight lines $\mathrm{x}+2\mathrm{y}-9=0,3\mathrm{x}+5\mathrm{y}-5=0$ and $\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y}-1=0$ are concurrent if the straight line $22\mathrm{x}-35\mathrm{y}-1=0$ passes through the point 

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. (a, b)

  2. (b,a)

  3. (-a,b)

  4. (-a, -b)

Show answer
Correct Option: B
Explanation:

$x+2y-9=0$ ---(1)


$3x+5y-5=0$ ---(2)


$ax+by-1=0$ ---(3)

Solving (1) and (2) simultaneously we get 

$y=22, x=-35$

Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $y=22, x=-35$ satisfy the third equation $ax+by-1=0$.

Substituting the values of 'x' and 'y' in this equation we get $-35a+22b-1=0$ ---(4)


And another equation given is $22x-35y-1=0$ ---(5)

Equation (5) will be of the form of equation (4), if we substitute

$x=b$ & $y=a$

That is $22x-35y-1=0$ passes through $(b,a)$

If $\mathrm{a}\neq b\neq \mathrm{c}$ and if $ax+by+\mathrm{c}=0\  bx+cy+\mathrm{a}=0$ and $cx+ay+b=0$ are concurrent, 

then find the value of 
$ 2^{\mathrm{a}^{2}b^{-1}\mathrm{c}^{-1}}2^{b^{2}\mathrm{c}^{-1}\mathrm{a}^{-1}}2^{\mathrm{c}^{2}\mathrm{a}^{-1}b^{-1}}$

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. 1

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: C

Which of the following are correct in respect of the system of equations $x + y + z = 8, x - y + 2z = 6$ and $3x - y + 5z = k$?
1. They have no solution, if $k = 15$.
2. They have infinitely many solutions, if $k = 20$.
3. They have unique solution, if $k = 25$.
Select the correct answer using the code given below

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3

Show answer
Correct Option: A
Explanation:

First we need to know if the above equations are linearly dependent or no,

In order to figure that out lets find the determinant $D$ which is;
$D=\left| \begin{matrix} 1 & 1 & 1 \ 1 & -1 & 2 \ 3 & -1 & 5 \end{matrix} \right| =0$
as $D=0$ the equations are linearly dependent, which means the system of equations can either be inconsistent or have infinitely many solutions. That shall depend on the value of k.
For there to be inifinitely many solutions , values of ${ D } _{ 1 }, { D } _{ 2 }, { D } _{ 3 }$ should be $0$ , defined as;
${ D } _{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \ 6 & -1 & 2 \ k & -1 & 5 \end{matrix} \right| \ { D } _{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \ 1 & 6 & 2 \ 3 & k & 5 \end{matrix} \right| \ { D } _{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \ 1 & -1 & 6 \ 3 & -1 & k \end{matrix} \right| $
which on solving you get;
${ D } _{ 1 }=\left| \begin{matrix} 8 & 1 & 1 \ 6 & -1 & 2 \ k & -1 & 5 \end{matrix} \right| =3k-60=3(k-20)\ { D } _{ 2 }=\left| \begin{matrix} 1 & 8 & 1 \ 1 & 6 & 2 \ 3 & k & 5 \end{matrix} \right| =k-20\ { D } _{ 3 }=\left| \begin{matrix} 1 & 1 & 8 \ 1 & -1 & 6 \ 3 & -1 & k \end{matrix} \right| =40-2k=-2(k-20)$
Now for all the Determinants to be zero , its evident that $k=20$
Hence statement 2 is correct.
Subsequently for there to be no solution $k\neq 20$, hence statement 1 is correct as well,
As $D$ is zero the system of equations CANNOT have a unique solution hence, statement 3 is WRONG.

To solve  $x + y = 3 : 3 x - 2 y - 4 = 0$  by determinant method find  $D.$

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $5$

  2. $1$

  3. $-5$

  4. $-1$

Show answer
Correct Option: A
Explanation:
${ a } _{ 1 }x+{ b } _{ 1 }y-{ c } _{ 1 }=0\quad \Rightarrow { a } _{ 1 }x+{ b } _{ 1 }y={ c } _{ 1 }$
${ a } _{ 2 }x+{ b } _{ 2 }y-{ c } _{ 2 }=0\quad \Rightarrow { a } _{ 2 }x+{ b } _{ 2 }y={ c } _{ 2 }$
then the solution of $x$ and $y$ can be obtained by evaluating the following integral :
$x=\frac { \left| \underset { { c } _{ 2 } }{ { c } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  }{ \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  } $  and  $y=\dfrac { \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { c } _{ 2 } }{ { c } _{ 1 } }  \right|  }{ \left| \underset { { a } _{ 2 } }{ { a } _{ 1 } } \quad \underset { { b } _{ 2 } }{ { b } _{ 1 } }  \right|  } $
$\therefore$    $x+y=3$
  $3x-2y=4$
can be solved using the above method
$x=\dfrac { \left| \underset { 4 }{ 3 } \quad \underset { -2 }{ 1 }  \right|  }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right|  } \quad ;\quad y=\dfrac { \left| \underset { 3 }{ 1 } \quad \underset { 4 }{ 3 }  \right|  }{ \left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right|  } $
$x=\dfrac { -6-4 }{ -2-3 } \quad ;\quad y=\dfrac { 4-9 }{ -5 } $
$x=\dfrac { 10 }{ -5 } \quad ;\quad y=\dfrac { -5 }{ -5 } $
$x=2\quad ;\quad y=1$
now the quantity $\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right| =D$ (determinant)
$D=\left| \underset { 3 }{ 1 } \quad \underset { -2 }{ 1 }  \right| =-5$
So, answer is option C.

If the lines $p _{1}x+q _{1}y=1,p _{2}x+q _{2}y=1 $ and $ p _{3}x+q _{3}y=1$ be concurrent, then the points $(p _{1},q _{1}),(p _{2},q _{2})$ and $(p _{3},q _{3})$ ,

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. are collinear

  2. form an equilateral triangle

  3. form a scalene triangle

  4. form a right angled triangle

Show answer
Correct Option: A
Explanation:

$p _{1}x+q _{1}y=1,\ p _{2}x+q _{2}y =1\ p _{3}x+q _{3}y=1$
Given lines are concurrent
$\Rightarrow \begin{vmatrix} p _{1} & q _{1} & 1 \ p _{2} & q _{2}& 1 \ p _{3} & q _{3} & 1 \end{vmatrix}=0$

$\Rightarrow p _{1}(q _{2}-q _{3})-q _{1}(p _{2}-p _{3})+(p _{2}q _{3}-p _{3}q _{2})=0$

$\Rightarrow (p _{1}q _{2}-p _{2}q _{1})+(p _{2}q _{3}-p _{3}q _{2})+(p _{3}q _{1}-p _{1}q _{3})=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates $(p _1, q _1),\; (p _2, q _2),\; (p _3,q _3)$
Since it is equal to zero, $(p _{1},q _{1}),(p _{2},q _{2}),(p _{3},q _{3})$ are collinear.

If $\Delta =\begin{vmatrix}
x+1 & x+2 & x+a\
x+2 & x+3 & x+b\
x+3 & x+4 & x+c
\end{vmatrix}=0$, then
the family of lines $ax+by+c=0$ passes through

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $(1, -1)$

  2. $(1, -2)$

  3. $(2, -3)$

  4. $(0, 0)$

Show answer
Correct Option: B
Explanation:

$\Delta =\begin{vmatrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{vmatrix}=0$


Applying ${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 },{ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$

$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \ x+2 & 1 & b-2 \ x+3 & 1 & c-3 \end{vmatrix}=0$

Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$

$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \ 1 & 0 & b-a-1 \ 2 & 0 & c-a-2 \end{vmatrix}=0$

Expanding along ${ C } _{ 2 }$
$1\left( c-a-2 \right) -2\left( b-a-1 \right) =0\ \Rightarrow a-2b+c=0$
From options for point $(1,-2)$ lies on the line $ax+by+c=0$

If the lines $\mathrm{x}+\mathrm{p}\mathrm{y}+\mathrm{p}=0,\ \mathrm{q}\mathrm{x}+\mathrm{y}+\mathrm{q}=0$ and $\mathrm{r}\mathrm{x}+\mathrm{r}\mathrm{y}+1 =0 (\mathrm{p},\mathrm{q}, \mathrm{r}$ being distinct and $ \neq$ 1) are concurrent, then the value of
$\displaystyle \frac{p}{p-1}+\frac{q}{q-1}+\frac{r}{r-1}=$

lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$

Show answer
Correct Option: A
Explanation:
$x+py+p=0$
$qx+y+q=0$
$rx+ry+1=0$
$\begin{vmatrix} 1 & p & p \\ q & 1 & q \\ r & r & 1 \end{vmatrix}$
$\left( 1-qr \right) -p\left( q-pqr \right) +p\left( qr-r \right) =0$
$1-qr-pq+pqr+pqr-pr=0$
$qr(p-1)+pr(q-1)-pq(r-1)-pqr=0$
$\Rightarrow \cfrac { p }{ p-1 } +\cfrac { q }{ q-1 } +\cfrac { r }{ r-1 } =1$
Option A
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