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Let a,b and c be three non-zero vectors such that no two of them are collinear and  (a xb)xc =   $\frac{1}{3}|b||c|a.$   . If θ is the angle between vectors b and c, then a value of   $\sin\theta$  is

Answer:

Explanation:

 Given,

   (a x b)x c=  $\frac{1}{3}|b||c|a.$

   $\Rightarrow$            -c x (a x b) = $\frac{1}{3}|b||c|a.$

     $\Rightarrow$      $-(c.b).a+(c.a)b= \frac{1}{3}|b||c|a$

             $\left[ \frac{1}{3}|b||c|+(c.b)\right]a=(c.a)b$

  Since a  and b are not collinear

          $c.b +\frac{1}{3}|b||c|=0 $   and c.a=0

  $\Rightarrow$       $|c||b|\cos \theta +\frac{1}{3}|b||c|=0 $

$\Rightarrow$      $|b||c|\left(\cos \theta +\frac{1}{3}\right)=0 $

$\Rightarrow$     $\cos \theta +\frac{1}{3}=0, (\therefore |b|\neq0,|c|\neq0)$

$\Rightarrow$      $\cos \theta =-\frac{1}{3}$

$\Rightarrow$      $\sin \theta =\frac{\sqrt{8}}{3}=\frac{2\sqrt{2}}{3}$

The equation of the plane containing the line 2x-5y+z=3, x+y+4z=5  and parallel  to the plane x+3y+6z=1 is

Answer:

Explanation:

 Let equation of plane  containing the line 2x-5y+z=3  and x+y+4z=5 be

   $(2x-5y+z-3)+\lambda (x+y+4z-5)=0$

$\Rightarrow  (2+\lambda)x+(\lambda-5)y+(4\lambda+1)z-3-5\lambda=0$      ........(i)

 This plane is parallel to the plane

 x+3y+6z=1

$\therefore$     $\frac{2+\lambda}{1}=\frac{\lambda-5}{3}=\frac{4\lambda+1}{6}$

 On taking first wo equalities , we get

                   $6+3\lambda=\lambda-5$

$\Rightarrow$                 $2\lambda=-11$

$\Rightarrow$            $\lambda=-\frac{11}{2}$

On taking last two equalities , we get 

                          $6\lambda-30=3+12\lambda$

$\Rightarrow$   $-6\lambda=33$

$\Rightarrow$     $\lambda=-\frac{11}{2}$

 So, the equation of required  plane is

  $\left(2-\frac{11}{2}\right)x+\left(\frac{-11}{2}-5\right) y+\left(-\frac{44}{2}+1\right) z-3+5\times \frac{11}{2}=0$

   $\Rightarrow$             $-\frac{7}{2}x-\frac{21}{2}y-\frac{42}{2}z+\frac{49}{2}=0$

   $\Rightarrow$   x+3y+6z-7=0

The distance of the point  (1,0,2)  from the point of intersection of the line  $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$   and the plane x-y+z=16 is

Answer:

Explanation:

 Given equation of the line is

                                  $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ =   $\lambda$              (say)......(i)

  and equation of plane is

  x-y+z=16        ...........(ii)

 Any point on the  line (i) is 

$(3\lambda+2,4\lambda-1,12\lambda+2)$

 Let this point be point of intersection of line and plane 

$\therefore$      $(3\lambda+2)-(4\lambda-1)+(12\lambda+2)=16$

 $\Rightarrow$        $11\lambda+5=16$

 $\Rightarrow$          $11\lambda=11$

$\Rightarrow$                       $\lambda=1$

  $\therefore$          Point  of intersection is (5,3,14).

 Now , distance  between  the points (1,0,2) and (5,3,14)

             $= \sqrt{(5-1)^{2}+(3-0)^{2}+(14-2)^{2}}$          

                  $=\sqrt{16+9+144}=\sqrt{169}=13$

 Let O be the vertex and Q be any point on the parabola   $x^{2}=8y$. If the point P divides the line segment OQ internally in the ratio 1:3, then the locus of P is

Answer:

Explanation:

 Central Idea.    Any point on the parabola   $x^{2}=8y$   is (4t,2t²) Point P divides the line segment joining of O(0,0) and

Q (4t,2t² ) in the ratio 1:3, Apply the section formula for internal division.

 Equation of parabola is   $x^{2}=8y$           .............(i)

 Ler any point  Q on the parabola (i) is   (4t, 2t²  )

 Let P (h,k)  be the point which divides the line segment joining (0,0) and ( 4t, 2t²)  in the ratio 1;3

$\therefore$   $h=\frac{1\times4t+3\times0}{4}$

$\Rightarrow $      $ h=t$

 and    $k=\frac{1\times 2t^{2}+3\times 0}{4d}$

  $\Rightarrow $         $k=\frac{t^{2}}{2}$

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

   $\Rightarrow $   2k=h²   $\Rightarrow $ 2y=  $x^{2}$, which is required locus             

The area (in sq units)  of the quadrilateral formed by the tangent at the endpoints of the latus rectum to  the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{5}=1$   is 

Answer:

Explanation:

 Given equation of ellipse is 

   $\frac{x^{2}}{5}+\frac{y^{2}}{5}=1$  

  $\therefore$     $a^{2}=9,b^{2}=5$  

$\Rightarrow$     a=3,  $b=\sqrt{5}$

Now,    $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$

   foci    $=  (\pm ae, 0)= (\pm 2,0)$    and    $\frac{b^{2}}{a}=\frac{5}{3}$

$\therefore$   Extremities of one latusrectum are

  $(2,\frac{5}{3})$

   $(2,\frac{-5}{3})$

 $\therefore$   Equation  of tangent at    $(2,\frac{5}{3})$  is

             $\frac{x(2)}{9}+\frac{y(5/3)}{5}=1$

  or 2x+3y= 9.......(ii)

   Eq .(ii)  intersects X and Y axes at   $(\frac{9}{2},0)$

     and (0,3) , respectively

  $\therefore$   Area of quadrilateral

                     =   $4\times  $  Area of   $\triangle POQ $

                      =  $4\times \left(\frac{1}{2} \times \frac{9}{2}\times 3\right)=27  sq.unit$

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