MHT CET 2020 :: Mathematics :: General questions

• Mathematics - General questions

 For any non-zero vectors a and b , [b  a x b a]=

Answer:

Explanation:

 We have

 [b a x b a]

 = [ a x b a b]            [ $\because$ [a b c]=[b c a]=[c a b]

 = (a x b) .(a x b) 

  = (a x b)²= | a x b|²

tan A+2tan 2A+4 tan 4A+8 cot 8A=

Answer:

Explanation:

 We have,

 tan A+ 2 tan 2A+4 tan 4 A+8 cot 8 A

 we know that

 cot A-tan A=2 cot 2 A

 $\therefore$           tan A= cot A-2cot 2A

 $\therefore$  cot A-2 cot 2A+2 tan 2A+4 tan 4 A+8 cot 8 A

 = cot A-  4 cot 4A+4 tan 4A+8 cot 8 A

  = cot A-8 cot 8 A+ 8 cot 8A = cot A

 If the line  $r= (\hat{i}-2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+2\hat{k})$    is parallel to the plane

$r. (3\hat{i}-2\hat{j}+m\hat{k})=10$ , then the value of m is 

Answer:

Explanation:

Given line, 

           $r= (\hat{i}-2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+2\hat{k})$    is parallel to the plane

$r. (3\hat{i}-2\hat{j}+m\hat{k})=0$

we know that, if line r= a+$\lambda$.b is parallel to plane  r.n=d

 then , b.n=0

 $\therefore$    $(2\hat{i}+\hat{j}+2\hat{k}) .(3\hat{i}-2\hat{j}+m\hat{k}) =0  $

 6-2+2m=0

    m=-2

 If  $A=\begin{bmatrix}2 & 3 \\1 & 2 \end{bmatrix}$  , $B=\begin{bmatrix}1 & 0 \\3 & 1\end{bmatrix}$ , then B^-1 A^-1 =

Answer:

Explanation:

We have 

      $A=\begin{bmatrix}2 & 3 \\1 & 2 \end{bmatrix}$ 

and    $B=\begin{bmatrix}1 & 0 \\3 & 1\end{bmatrix}$

$AB=\begin{bmatrix}2 & 3 \\1 & 2 \end{bmatrix}  \begin{bmatrix}1 & 0 \\3 & 1\end{bmatrix} $

$AB=\begin{bmatrix}11 & 3 \\7 & 2 \end{bmatrix}  $

 |AB|=22-21=1

$\therefore$    $(AB)^{-1}=\begin{bmatrix}2 & -3 \\-7 & 11 \end{bmatrix} $

 $\therefore$     We know that 

$B^{-1}A^{-1}=(AB)^{-1}$

$\therefore$     $B^{-1}A^{-1}=\begin{bmatrix}2 & -3 \\-7 & 11 \end{bmatrix}  $

The angle between the lines   $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$  and

$\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$

Answer:

Explanation:

Given line,

$\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$  and

$\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$

 Here, a₁=4, b₁=1, c₁=8

 a₂=2, b₂ =2, c₂=1

$\therefore$    $\cos\theta=\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{(\sqrt{a_1^2+b_1^2+c_1^2})(\sqrt{a_2^2+b_2^2+c_2^2})}$

$\Rightarrow $        $\cos\theta=\frac{8+2+8}{(\sqrt{16+1+64})(\sqrt{4+4+1})}$

$\Rightarrow $       $\cos\theta=\frac{18}{9\times3}=\frac{2}{3}$

 $\Rightarrow $     $ \theta=\cos^{-1}\left(\frac{2}{3}\right)$

• Mathematics - General questions