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 If   $[a\times b .b\times c. c\times a]=\lambda[abc]^{2}$  then λ  is equal to

Answer:

Explanation:

 Use the following formula to simplify

   a x (bx c)=(a.c)b-(a.b)c

               [abc]=[bca]=[cab]

 and [a a b]=[a b b]=[a c c]=0

Now, [ a x ba x c c x a]

      = a xb.(( b xc ) x (c xa))

      = a x b.(k x (c x a))

 Let k= b x c

        = a x  b.((k.a) c-(k.c).a)

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

  = ( a x b). ([b c a ] c)-0

                               [ $\because$   [  b x c.c ]=0]

 = (a x b).c [ b c a]

  = [ a b c ] [ b c a]= [ a b c]²

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

 But given,

  [ a x b b x c  c xa]= λ  [ a b c]²

So  [a b c]²  = λ  [ a b c ]²

$\Rightarrow$                 $\lambda$  =1

The image of the line   $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$    in the plane 2x-y+z+3=0 is the line

Answer:

Explanation:

Here, the plane, line and its image are parallel to each other. So, find any point on the normal to the plane from which the image line will be passed and then find the equation of the image line.

 Here, plane and line are parallel to each other, Equation of normal to the plane through the point (1,3,4) is 

   $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}$ = k              [say]

 Any point in this normal is 

                         (2k+1, -k+3,4+k)

  Then   $\left(\frac{2k+1+1}{2},\frac{3-k+3}{2},\frac{4+k+4}{2}\right)$   lies on plane  $\Rightarrow 2$

  $(k+1)-\left(\frac{6-k}{2}\right)+\left(\frac{8+k}{2}\right)+3=0$ 

    $\Rightarrow$               k=- 2

 Hence, point through  which this image pass is 

       (2k+1, 3-k,4+k)

 i.e,     [2(-2)+1.3+2.4-2]=(-3.5,2)

 Hence , equation of image line is

                                   $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

 The slope of the line touching both the parabola   $y^{2}=4x$    and $x^{2}=-32y$ is 

Answer:

Explanation:

 Let the tangent  to parabola be y=mx+a/m, If it touches the other  curve, then D=0, to get the value of m 

 For parabola, $y^{2}=4x$ 

  Let y=mx+1/m  be tangent line and it touches the parabola  $x^{2}=-32y$

   $\therefore$   $ x^{2}=-32\left(mx+\frac{1}{m}\right)$

$\Rightarrow $     $x^{2}+32mx+\frac{32}{m}=0$

$\therefore$     D=0

$\therefore$    $(32m)^{2}-4.\left(\frac{32}{m}\right)=0$

    $\Rightarrow $                    $m^{3}=\frac{1}{8}$

      $\therefore$                      m$=\frac{1}{2}$

 Let C be the circle with centre at (1,1) and radius 1. If T is the circle centred at (0,y)  passing through the origin and touching the circle C externally, then the radius of T is equal to 

Answer:

Explanation:

 Use the property, when two circles touch each other externally, then distance between the centre is equal to sum of their radii, to get the required radius.

 Let the coordinate of the centre of T be (0,k)

 Distance between their centre

   $k+1=\sqrt{1+(k-1)^{2}}  $           [  $\because $   C₁C2=k+1]

$\Rightarrow$      $ k+1= \sqrt{1+k^{2}+1-2k} $

$\Rightarrow$        $ k+1= \sqrt{k^{2}+2-2k} $

$\Rightarrow$        $ k^{2}+1+2k= k^{2}+2-2k $

$\Rightarrow$       k=1/4

So  , the radius of circle T is  k, i.e, 1/4

 The locus of the foot of perpendicular drawn from the centre of the ellipse    $x^{2}+3y^{2}=6 $ on any tangent to it is

Answer:

Explanation:

Equation of ellipse is   $x^{2}+3y^{2}=6 $   or $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$

  Equation of the tangent is 

$\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$

Let (h,k) be any point on the locus.

$\therefore$     $\frac{h}{a}\cos\theta+\frac{k}{b}\sin\theta=1$      .....(i)

 Slope  of the tangent   $\frac{-b}{a}\cot\theta$

  Slope of perpendicular drawn from centre (0,0) to (h,k) is k/h

Since, both the lines are perpendicular

$\therefore$    $\left(\frac{k}{h}\right)\times(\frac{-b}{a}\cot\theta)=-1$

    $\Rightarrow$      $\frac{\cos\theta}{ha}=\frac{\sin\theta}{kb}=\alpha$          [say]

$\Rightarrow$     $\cos\theta=\alpha ha$

                      $\sin \theta=\alpha kb$

From Eq(i) , we get

     $\frac{h}{\alpha}(\alpha ha+\frac{k}{b}(\alpha kb)=1$

 $\Rightarrow$      $h^{2}\alpha+k^{2}\alpha=1$

$\Rightarrow$             $\alpha=\frac{1}{h^{2}+k^{2}}$

Also,           $\sin^{2}\theta+\cos^{2}\theta=1$

    $\Rightarrow$           $(\alpha kb)^{2}+(\alpha ha)^{2}=1$

$\Rightarrow$        $\alpha^{2}k^{2}h^{2}+\alpha^{2}h^{2}a^{2}=1$

  $\Rightarrow$                              $\frac{k^{2}b^{2}}{(h^{2}+k^{2})^{2}}+\frac{h^{2}a^{2}}{(h^{2}+k^{2})^{2}}=1$

    $\Rightarrow$                        $\frac{2k^{2}}{(h^{2}+k^{2})^{2}}+\frac{6h^{2}}{(h^{2}+k^{2})^{2}}=1$

                                                                                                   [$\therefore$  a²=6, b²=2]

$\Rightarrow$        $6x^{2}+2y^{2}=(x^{2}+y^{2})^{2}$

                                      [Replaceing k by y and h by x]

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