Q: If | A \times B | = \sqrt3 A.B , then the value of | A + B | is :

We know that, A \times B = AB \sin \theta Also, A.B = AB \cos \theta Given : | A \times B | = \sqrt3 A.B Using | A \times B | = |A| |B| \sin \theta We get |A\times B| = AB \sin \theta A.B = |A||B| \cos \theta \therefore AB \sin \theta = \sqrt3 AB \cos \theta \tan \theta = \sqrt3 \implies \theta = 60^o Now (A+B)^2 = A^2 + B^2 + 2 A.B = A^2 + B^2 + 2 AB \cos\theta = A^2 + B^2 + 2AB\times \dfrac {1}{2} = A^2 + B^2 + AB or |A+B| = ( A^2 + B^2 + AB )^{1/2}

Question 1

If   | A 	imes B | = \sqrt3 A.B ,   then the value of   | A + B |   is :

Accepted Answer

We know that,  A 	imes B = AB \sin 	heta  Also,    A.B = AB \cos 	heta  Given :    | A 	imes B | = \sqrt3 A.B  Using    | A 	imes B | = |A| |B| \sin 	heta  We get   |A	imes B| = AB \sin 	heta   A.B = |A||B| \cos 	heta  	herefore     AB \sin 	heta = \sqrt3 AB \cos 	heta   	an 	heta = \sqrt3   \implies   	heta = 60^o  Now   (A+B)^2 = A^2 + B^2 + 2 A.B   = A^2 + B^2 + 2 AB \cos	heta  = A^2 + B^2 + 2AB	imes  \dfrac {1}{2}   = A^2 + B^2 + AB  or   |A+B| = ( A^2 + B^2 + AB )^{1/2}

Question 2

A line makes angles  \alpha,\beta,\gamma,\delta  with the four diagonals of a cube then  \cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta  is equal to

Accepted Answer

REF.ImageTake 'O' as a corner OA,OB,OC  are 3 edges through the axesLet  OA=OB=OC=a coordinates of  O=(o,o,o) A(a,o,o) B(o,a,o) C(o,o,a) P(a,a,o) L(o,a,a) M(a,o,a) N(a,a,o) The four diagonals  OP, AL, BM, CN Direction cosine of  OP : a-o,a-o,a-o = a,a,a = 1,1,1 Direction cosine of  AL : o-a,a-o,a-o = -a,a,a = -1,1,1 Direction cosine of  BM : a-o,o-a,a-o = a,-a,a = 1,-1,1 Direction cosine of  CN : a-o,a-o,o-a = a,a,-a = 1,1,-1 	herefore   DC's of OP are  \displaystyle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} DC's of AL are  \displaystyle\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} DC's of BM are  \displaystyle\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}} DC's of CN are  \dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}} Let l,m,n be dc's of line and line makes angle  \alpha  with OP :-  \displaystyle \cos \alpha = l(\frac{1}{\sqrt{3}})+m(\frac{1}{\sqrt{3}})+n(\frac{1}{\sqrt{3}})=\frac{l+m+n}{\sqrt{3}} Similarly  \cos\, \beta =\dfrac{-l+m+n}{\sqrt{3}}   \cos \delta  = \dfrac{l+m-n}{\sqrt{3}} \cos\gamma =\dfrac{l-m+n}{\sqrt{3}} suaring and adding all the fouri.e ;  \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta  = \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}] =\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2}) [\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3} 	herefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}

Question 3

Find the vector equation of the passing through the point  2i+3j+k  and parallel to the vector  4i-2j+3k .

Accepted Answer

The line passes through the point,

\overrightarrow a 
= 2\hat i + 3\hat j + \hat k

And parallel to the line,

\overrightarrow b 
= 4\hat i - 2\hat j + 3\hat k

The vector equation of the line passing through a
point and parallel to the given line is,

\overrightarrow r 
= \overrightarrow a  + \lambda
\overrightarrow b

Substituting the values, we get,

\overrightarrow r 
= \left( {2\hat i + 3\hat j + \hat k} \right) + \lambda \left( {4\hat i
- 2\hat j + 3\hat k} \right)

\overrightarrow r 
= \left( {2 + 4\lambda } \right)\hat i + \left( {3 - 2\lambda }
\right)\hat j + \left( {1 + 3\lambda } \right)\hat k

Question 4

The distance of the point (1,-2,3) from the plane x-y+z=5, measured parallel to the line  \dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-6}

Accepted Answer

Point is  (1, -2, 3) Equation of plane =  x-y+z=5 Equation of parallel line is  \dfrac{x}{2} = \dfrac{y}{3} =\dfrac{z}{-6} \dfrac{x-1}{2} = \dfrac{y+2}{3} = \dfrac{z-3}{-6} = \lambda x=2\lambda + 1 , y=3\lambda-2 , z=-6\lambda+3 Since, x-y+z=5 2\lambda+1-3\lambda+2-6\lambda+3=5 -7\lambda+6=5 \lambda=\dfrac{1}{7} Thus, x=\dfrac{9}{7} , y=\dfrac{-11}{7} , z=\dfrac{15}{7} Therefore,Distance =  \sqrt{(\dfrac{2}{7})^{2} + (\dfrac{3}{7})^{2} + (\dfrac{-6}{7})^{2}}  =  \sqrt{\dfrac{49}{49}}  =  1  unit

Question 5

The equation of the plane containing the two lines  \displaystyle \frac { x-1 }{ 2 } =\frac { y+1 }{ -1 } =\frac { z }{ 3 }   and  \displaystyle \frac { x }{ 2 } =\frac { y-2 }{ -1 } =\frac { z+1 }{ 3 }   is

Accepted Answer

Equation of any plane through the first line is  a\left( x-1 \right) +b\left( y+1 \right) +cz=0     ...(1)where  2a-b+3c=0     ...(2)It will pass through the second line if  (0,2,-1),  a point on the second line lies on it i.e., if  -a+3b-c=0      ...(3)Solving (2) and (3), we get  \displaystyle \frac { a }{ -8 } =\frac { b }{ -1 } =\frac { c }{ 5 }   or  \displaystyle \frac { a }{ 8 } =\frac { b }{ 1 } =\frac { c }{ -5 }  Hence, equation of required plane is  8\left( x-1 \right) +\left( y+1 \right) -5z=0 \Rightarrow 8x+y-5z-7=0.

Question 6

Find the equation of the plane passing through the following points: (2,3,4), (-3, 5, 1)  and  (4,-1,2)

Accepted Answer

Vector equation of a plane passing through three points with position vectors \vec{a},\vec{b},\vec{c} is \left(\vec{r}-\vec{a} ight).\left[\left(\vec{b}-\vec{a} ight) imes\left(\vec{c}-\vec{a} ight) ight]=0 Now, the plane passes through the points A\left(2,3,4 ight),\,B\left(-3,5,1 ight) and C\left(4,-1,2 ight) \vec{b}-\vec{a}=-3\hat{i}+5\hat{j}+\hat{k}-2\hat{i}-3\hat{j}-4\hat{k}=-5\hat{i}+2\hat{j}-3\hat{k} \vec{c}-\vec{a}=4\hat{i}-\hat{j}+2\hat{k}-2\hat{i}-3\hat{j}-4\hat{k}=2\hat{i}-4\hat{j}-2\hat{k} \left(\vec{b}-\vec{a} ight) imes\left(\vec{c}-\vec{a} ight)=\left|

\hat{i} - 5 2 \hat{j} 2 - 4 \hat{k} - 3 - 2

ight| =\left(-4-12 ight)\hat{i}-\left(10+6 ight)\hat{j}+\left(20-4 ight)\hat{k} =-16\hat{i}-16\hat{j}+16\hat{k} herefore\, vector equation of plane is \left(x\hat{i}+y\hat{j}+z\hat{k}-2\hat{i}-3\hat{j}-4\hat{k} ight)\left(8\hat{i}-16\hat{j}-16\hat{k} ight)=0 \Rightarrow\,\left[\left(x-2 ight)\hat{i}+\left(y-3 ight)\hat{j}+\left(z-4 ight)\hat{k} ight]\left(8\hat{i}-16\hat{j}-16\hat{k} ight)=0 \Rightarrow\,8x-16-16y+48-16z+64=0 \Rightarrow\,8x-16y-16z+96=0 or x-2y-2z+12=0 is the required equation of the plane.

Question 7

The image of the point  (1,3,4)  in the plane  2x -y+z+3=0  is

Accepted Answer

Let   P (1,3,4)   be a point Let M be the point on the planeEquation of the plane is   2x-y+z+3 = 0  Thus equation of the plane is;  \dfrac{x-1}{2} = \dfrac{y-3}{-1} = \dfrac{z-4}{1} = k  Any point on the above line, PM is of the form  x = 2k+1;y = -k+3; z = k+4  Substituting the above value in the equationof the place we have  2(2k+1)-(-k+3)+(k+4)+3 = 0   \Rightarrow 4k+2+k-3+k+4+3 = 0   \Rightarrow 6k+6 = 0   k = -1  Thus coordinates on M are  x = 2(-1)+1 = -1   y =-(-1)+3 = 4   z = (-1)+4 = 3  Let   Q (x_{1},y_{1},z_{1})   be the image of PThus M is the midpoint of PQ Therefore,  -1 = \dfrac{1+x_{1}}{2}; 4 = \dfrac{3+y_{1}}{2}; 3 = \dfrac{4+z_{1}}{2}   \Rightarrow x_{1} = -3, y_{1} = 5, z_{1} = 2

Question 8

If a line makes angles  \alpha, \beta, \gamma  with the coordinate axes, then the value of  \cos 2\alpha + \cos 2\beta + \cos 2\gamma  is

Accepted Answer

\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2(\cos^{2} \alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3 = 2 - 3 = -1 (\because \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1)

Question 9

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Accepted Answer

Let the direction cosines of the line make an angle  a  with each of the coordinate axes. 	herefore   l=\cos {a}, m=\cos {a}, n=\cos {a} We know  {l}^{2}+{m}^{2}+{n}^{2}=1 \Rightarrow   \cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \alpha  } =1 \Rightarrow   3\cos ^{ 2 }{ \alpha  } =1 \Rightarrow    \cos ^{ 2 }{ \alpha  } =\cfrac{1}{3} \Rightarrow   \cos {\alpha}=\pm \cfrac{1}{\sqrt 3} Thus, the direction cosines of the line which is equally inclined to the coordinate axes are  \pm \cfrac{1}{\sqrt 3}, \pm \cfrac{1}{\sqrt 3}   and  \pm \cfrac{1}{\sqrt 3} .

Question 10

Find shortest distance between lines  \vec r=\hat i+2\hat j+3\hat k+\lambda (2 \hat i+\hat j+4\hat k)  and \vec r=2\hat i+4 \hat j+5\hat k+\mu( 3\hat i+4\hat j+5\hat k)

Accepted Answer

Given eq of lines \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k}) ----(1) \vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu(3\hat{i}+4\hat{j}+5\hat{k}) ----(2)position vector and normal vector of line (1) \vec{a}=\hat{i}+2\hat{j}+3\hat{k} \vec{n_{1}}=2\hat{i}+3\hat{j}+4\hat{k} position vector and normal vector of line (2) \vec{c}=2\hat{i}+4\hat{j}+5\hat{k} \vec{n_{2}}=3\hat{i}+4\hat{j}+5\hat{k} shortest between two skews line SD=\dfrac{\vec{AC}\cdot(\vec{n_{1}} imes\vec{n_{2}})}{\left | \vec{n_{1}} imes\vec{n_{2}} ight |} AC=\vec{c}-\vec{a} AC=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k}) AC=2\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k} AC=\hat{i}+2\hat{j}+2\hat{k} \vec{n_{1}} imes\vec{n_{2}}=

\hat{i} - 5 2 \hat{j} 2 - 4 \hat{k} - 3 - 2

\vec{n_{1}} imes\vec{n_{2}}=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9) \vec{n_{1}} imes\vec{n_{2}}=-\hat{i}+2\hat{j}-\hat{k} \left | \vec{n_{1}} imes\vec{n_{2}} ight |=\sqrt{(-1)^2+2^2+(-1)^2} \left | \vec{n_{1}} imes\vec{n_{2}} ight |=\sqrt{6} putting AC \vec{n_{1}} imes\vec{n_{2}},\left | \vec{n_{1}} imes\vec{n_{2}} ight | in formula SD=\dfrac{\vec{AC}\cdot(\vec{n_{1}} imes\vec{n_{2}})}{\left | \vec{n_{1}} imes\vec{n_{2}} ight |} SD=\dfrac{(\hat{i}+2\hat{j}+2\hat{k})\cdot(-\hat{i}+2\hat{j}-\hat{k})}{\sqrt{6}} SD=\dfrac{-1+4-2}{\sqrt{6}} SD=\dfrac{1}{\sqrt{6}}