VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

The volume of the tetrahedron included between the plane 3x+4y-5z-60=0  and the coordinate planes is 

Answer:

Explanation:

The given equation of the plane is $3x+4y-5z-60=0$ it can be written in the form $\frac{x}{20}+\frac{y}{15}+\frac{z}{-12}=1$

 which meets the coordinate axes at the pints A(20,0,0), B(0,15,0) and C(0,0,-12) .The coordinates of the origin are O(0,0,0) Therefore , volume of the tetrahedron  OABC is = 

$\frac{1}{6}\begin{bmatrix}20 &0&0 \\0 & 15&0\\0&0&-12 \end{bmatrix}=\frac{1}{6}|20\times15\times(-12)|=600$

if a plane meets the coordinate axes at A, B and C  such that the centroid of the triangles is (1,2,4) , then the equation of te plane is 

Answer:

Explanation:

Let the equation of the plane is 

$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1$

Then, A($\alpha$,0,0) , B(0,$\beta$, 0),and C(0,0,$\gamma$) are the points on the coordinates axes, The centroid of the triangle is (1,2,4)

$\therefore$     $\frac{\alpha}{3}=1 \Rightarrow \alpha=3$

$\frac{\beta}{3}=2 \Rightarrow  \beta=6$

and $\frac{\gamma}{3}=2 \Rightarrow  \gamma=12$

$\therefore$  The equation of the plane is 

  $\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1$

$\Rightarrow$   $4x+2y+z=12$

Equation of the chord of the hyperbola $25x^{2}-16y^{2}=400$ which is bisected at the point (6,2) is 

Answer:

Explanation:

Given hyperbola is $25x^{2}-16y^{2}=400$

 If (6,2) is the mid point of the chord , then equation of chord is $T=S_{1}$

 $\Rightarrow$       $25(6x)-16(2y)=25(36)-16(4)$

$\Rightarrow$    $75x-16y=450-32$

$\Rightarrow$      $75x-16y=418$

If the area of the triangle on the complex plane formed by the points z, $z+iz$ and iz is 200 , then the value of 3|z| must be equal to 

Answer:

Explanation:

 Let $z=x+iy$, then

 $z+iz=x+iy+i(x+iy)=(x-y)+i(x+y)$

and $iz=i(x+iy)=-y+ix$

 Then, the area of the triangle formed by these lines is 

   $\triangle$=$\frac{1}{2} \begin{bmatrix}x &y&1 \\(x-y) & (x+y)&1\\-y&x&1 \end{bmatrix}$

 Applying  $R_{2} \rightarrow  R_{2}-(R_{1}+R_{3})$

 $\triangle$=$\frac{1}{2} \begin{bmatrix}x &y&1 \\0& 0&-1\\-y&x&1 \end{bmatrix}$=$\frac{1}{2}(x^{2}+y^{2})$

$\Rightarrow$   $\frac{1}{2}|z|^{2}=200$ (given)

 $|z|^{2}=400 \Rightarrow |z|=20$

 $\therefore$  $3|z|=3 \times 20=60$

The angle between lines joining the origin to the point  of intersection of the line $\sqrt{3} x+y=2$ and the the curve $y^{2}-x^{2}=4$ is 

Answer:

Explanation:

On homogenising $y^{2}-x^{2}=4$  with the help of the line $\sqrt{3}x+y=2$, we get

$y^{2}-x^{2}=4\frac{\left(\sqrt{3}x+y\right)^{2}}{4}$

$\Rightarrow$      $y^{2}-x^{2}=3x^{2}+y^{2}+2\sqrt{3}xy$

 $\Rightarrow$  $4x^{2}+2\sqrt{3} xy=0$

On comparing with $ax^{2}+2hxy+by^{2}=0$ .

we get,

  $a=4,h=\sqrt{3}$ and $b=0$

We know that,

$\tan \theta= 2\frac{\sqrt{h^{2}-ab}}{a+b}=\frac{2\sqrt{3-0}}{4+0}$

$\therefore$ The angle between the lines is 

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

• Mathematics - General questions