The direction ratios of normal to the plane through the points (0,-1,0) and (0,0,1) and making an angle \frac{\pi}{4}  with the plane y-z+5=0 are :

  • Option 1)

     

    \sqrt2,1,-1

  • Option 2)

    2,-1,1

  • Option 3)

     

    2,\sqrt2,-\sqrt2

  • Option 4)

     

    2\sqrt3,1,-1

Answers (1)
A admin

 

Cartesian equation of plane passing through a given point and normal to a given vector -

\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0
 

- wherein

\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}

\vec{a}= x_{0}\hat{i}+y_{0}\hat{j}+z_{0}\hat{k}

\vec{n}= a\hat{i}+b\hat{j}+c\hat{k}

Putting in

\left ( \vec{r}-\vec{a} \right )\cdot \vec{n}= 0

We get \left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0

 

 

Angle between two planes (Cartesian form) -

Let the two planes be

ax+by+cz+d=0\: and \: a_{1}x+b_{1}y+c_{1}z+d_{1}=0

then the angle between them is defined as the angle between their normals

\cos \Theta = \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}

-

 

 

Equation of line passing through \left ( 0,-1,0 \right ) & \left ( 0,0,1 \right )  is   \frac{x-0}{0}=\frac{y+1}{1}=\frac{z-0}{1}=\lambda

Now, \left | \frac{\bar{n_{1}}\cdot \cdot \bar{n_{2}}}{\left | n_{1} \right |\cdot \left | n_{2} \right |} \right |=\cos \frac{\pi }{4}

\left | \frac{-1-1}{\sqrt{2}\sqrt{a^{2}+2}} \right |=\frac{1}{\sqrt{2}}

2=\sqrt{a^{2}+2}\Rightarrow a^{2}+2=4\Rightarrow a=\pm \sqrt{2}

So direction ratios : \left ( \sqrt{2},-1,1 \right )or\left ( -\sqrt{2} ,-1,1\right )

So answer are (1) and (3)


Option 1)

 

\sqrt2,1,-1

Option 2)

2,-1,1

Option 3)

 

2,\sqrt2,-\sqrt2

Option 4)

 

2\sqrt3,1,-1

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