Question

$\begin{array}{cccc}& \text{List-1}& & \text{List-2}\\ \left(P\right)& \text{Let L be the line}\frac{x-6}{2}=\frac{y-12}{4}=\frac{z-8}{2}& & 1.5\\ & \text{Let A be the point of intersection of the}& & \\ & \text{plane}{p}_1:x+y+z=6\text{and}L.\text{Let P be the}& & \\ & \text{point of intersection of the plane}{p}_2:x+y+z=10& & 2.8\\ & \text{L. AP is produced to Q such that AP=PQ}& & \\ & \text{and if coordinates of Q are}(\alpha ,p,r)then& & 3.6\\ & \text{distnce of Q from xy-plane is}& & \\ \left(q\right)& \text{DABC is a tetrahedran A(2,0,0), B(0,4,0})& & 4.2\\ & \text{Edge CD lies on the line}\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}.\text{If}& & 5.0\\ & \text{locus of centroid of tetrahedron lies on}& & \\ & \frac{x-x_1}{1}=\frac{y-y_1}{a}=\frac{z-z_1}{b}\text{then a+b is}& & \\ \left(R\right)& \text{P is the area of region of complex plane}& & \\ & \text{generated by}(z:\frac{z}{4}\text{and}\frac{4}{\overline{z}}\text{have real and}& & \\ & \text{imaginary part in (0,1) )then [p]is}& & \\ & \text{where [,] denotes greatest integer function}& & \\ \left(S\right)& \text{If}{Z}_1=a_1+i{b}_{1}and{z}_2=a_2+i{b}_2\text{are complex}& & \\ & \text{numbers such that}|z_1|=1,|z_2|=2\text{and}& & \\ & Re\left(z_1{z}_2\right)=0\text{and let}{w}_1=a_1+\frac{i{a}_2}{2}\text{and}& & \\ & w_2=2b_1+i{b}_2\text{then}|\left(w_1{w}_2\right)|is& & \end{array}$

• A. $p\to 1,q\to 3,r\to 4,s\to 5$
• B. $p\to 1,q\to 1,r\to 3,s\to 5$
  
• C. $p\to 1,q\to 1,r\to 1,s\to 5$
  
• D. $p\to 4,q\to 1,r\to 1,s\to 4$

Answer 1

The correct option is C $p\to 1,q\to 1,r\to 1,s\to 5$


$\left(P\right)\frac{x-6}{2}=\frac{y-12}{4}=\frac{z-8}{2}$

$(2\lambda +6,4\lambda +12,2\lambda +8)$

$P_1\to x+y+z=6$

$\lambda =-\frac{5}{2}$

$A(1,2,3)$

$P_2\to x+y+z=10\Rightarrow \lambda =-2$

$P(2,4,4)$



(Q) $A(2,0,0),B(0,4,0)C(\alpha +1,2\alpha +2,3\alpha +3)$

$D(\beta +1,2\beta +2,3\beta +3)$

Centroid is $(\frac{\alpha +\beta +4}{4},\frac{2(\alpha +\beta )+8}{4},\frac{3(\alpha +\beta )+6}{4})$

Required locus is $4x-4=\frac{4y-8}{2}=\frac{4z-6}{3}$

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

(R)

Area of Region = $16-(2\pi +2\pi -(2\pi -4\left)\right)=12-2\pi$

(s) $a_1^2+b_1^2=1\dots \left(1\right)$,

$a_2^2+b_2^2=4\dots \left(2\right)$

$a_1{a}_2=b_1{b}_2$

$\left(1\right)\times \left(2\right)\Rightarrow a_1^2{a}_2^2+b_1^2{b}_2^2+a_1^2{b}_2$

$\Rightarrow (a_1{b}_2+a_2{b}_1{)}^2=4$

$\Rightarrow a_1{b}_2+a_2{b}_1=\pm 2$