VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

If a plane  passing through the point (2,2,1) and is perpendicular to the planes 3x+2y+4z+1=0 and  2x+y+3z+2=0 . Then the equation of the plane is 

Answer:

Explanation:

Equation of plane passing through (2,2,1) is

 a(x-2) +b(y-2)+c(z-1)=0   .....(i)

 Since , above plane is perpendicular to 

3x+2y+4z+1=0

 and 2x+y+3z+2=0

$\therefore$    3a+2b+4c=0       ......(ii)

 and 2a+b+3c=0     .....(iii)

 [ $\because$ for perpendicular , a₁a₂+b₁b₂+c₁c₂=0]

On multiplying eq.(iii) by 2 , we get

 4a+2b+6c=0     ......(iv)

 On subtracting eq.(iv) from eq.(ii) , we get

$\Rightarrow  c=\frac{-a}{2}$

 On putting $  c=\frac{-a}{2}$ in eq. (iii) , we get b = $\frac{-a}{2}$

 on putting b = $\frac{-a}{2}$ and $c=\frac{-a}{2}$ in eq.(i)

we get   $a(x-2)-\frac{a}{2}(y-2)-\frac{a}{2}(z-1)=0$

$\Rightarrow\frac{a}{2}[2(x-2)-(y-2)-(z-1)]=0$

$\Rightarrow2x-4+y+2-z+1=0$

$\Rightarrow2x-y-z-1=0$

The shortest distance between the lines  $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$  and 

$\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}$ is 

Answer:

Explanation:

 Given, lines are   

$\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$  and 

$\frac{x-10}{3}=\frac{y-304}{8}=\frac{4-z}{5}$ 

 The vector  form of given lines are

  $r= 7\hat{i}-4\hat{j}+6\hat{k} +\lambda (3\hat{i}-16\hat{j}+7\hat{k})$   and

$r=10\hat{i}+30\hat{j}+4\hat{k} +\mu (3\hat{i}+8\hat{j}-5\hat{k})$

 On comparing these equations with

$r=a_{1}+\lambda b_{1}$  and   $r=a_{2}+\mu b_{2}$, we get 

$\overrightarrow{a_{1}}=7\hat{i}-4\hat{j}+6\hat{k} $

$\overrightarrow{a_{2}}=10\hat{i}+30\hat{j}+4\hat{k} $

$\overrightarrow{b_{1}}=3\hat{i}-16\hat{j}+7\hat{k} $

 and $\overrightarrow{b_{2}}=3\hat{i}+8\hat{j}-5\hat{k} $

 Shortest distance=    $|\frac{(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}).(\overrightarrow{b_{1}}\times\overrightarrow{b_{2}})}{|\overrightarrow{b_{1}}\times\overrightarrow{b_{2}}|}|$

 = $|\frac{(3\hat{i}+34\hat{j}-2\hat{k}).(24\hat{i}+36\hat{j}+72\hat{k})}{84}|$

 = $|\frac{72+1224-144}{84}|=|\frac{1152}{84}|=\frac{288}{41}$  units

 If the mean and variance of a binomial  distribution are 4 and 2, respectively . Then the probability of atleast 7 successes is 

Answer:

Explanation:

 Here, mean =4 and variance =2

 $\Rightarrow$   np=4 and npq=2

 so,  $\frac{npq}{np}=\frac{2}{4}\Rightarrow q=\frac{1}{2}$

 Then   $p=1-q=1-\frac{1}{2}=\frac{1}{2}$

 Mean= np=4

 $\Rightarrow $   $n\times\frac{1}{2}=4\Rightarrow n=8$

 $\therefore$      $P(X=r)=^{n}C_{r} p^{r}q^{n-r}$

 $=^{8}C_{r} \left(\frac{1}{2}\right)^{8}$       $  \left[\because p=q=\frac{1}{2}\right]$

 The required probabilty of atleast 7 successes is

  $P(X\geq 7)=P(X=7)+P(X=8)$

 = $\left(^{8}C_{7}+^{8}C_{8}\right)\left(\frac{1}{2}\right)^{8}$

  =   $\left(\frac{8!}{7!1!}+\frac{8!}{8!0!}\right)\left(\frac{1}{2}\right)^{8}$

=  $\left(8+1\right)\left(\frac{1}{2}\right)^{8}=\frac{9}{256}$

$\int_{0}^{\pi/2} \sin 2x.\log \tan x dx$ is equal to

Answer:

Explanation:

Let   $I=\int_{0}^{\pi/2} (\log \tan x).\sin 2 x dx$   ......(i)

$I=\int_{0}^{\pi/2} \log \tan(\frac{\pi}{2}- x).\sin 2(\frac{\pi}{2}- x) dx$

                                           $\left[\because\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx\right]$

$\Rightarrow I=\int_{0}^{\pi/2} \log \cot x.\sin 2 x dx$...........(ii)

                                         $[\because\sin (\pi-2x)=\sin 2x]$

 On adding eqs(i) and (ii) , we get

$2I\int_{0}^{\pi/2} \log\tan x.\sin2x dx+\int_{0}^{\pi/2} \log\cot x\sin 2x dx$

=  $\int_{0}^{\pi/2} \sin2x\log(\tan x.\cot x) dx$

$[\because$   $ \log m+\log n=\log (m.n)]$

 $\int_{0}^{\pi/2} \sin 2x\log1 dx$

$\Rightarrow$   $1=0[\because log 1=0]$

$\therefore $   $\int_{0}^{\pi/2} \sin 2x\log (\tan x) dx=0$

  For any three vectors a,b and c , [a+b, b+c, c+a] is

Answer:

Explanation:

 [a+b , b+c, c+a]

= (a+b).[(b+c) x (c+a)]

= (a+b).[b x c +b x a+c x c+ c x a]

= (a+b).(b x c +b x a+ c x a)

                            [$\because$ c x c=0]

 = a.(b x c)+ a.(b x a)+a. (c x a)+ b. (bx c)+b. (b x a)+b.(c x a)

 =a.(b x c)+b. (c x a)

 =[a b c]+[a b c]

=2[a b c]

• Mathematics - General questions