CBSE Questions for Class 6 Maths Basic Geometrical Ideas Quiz 7 - MCQExams.com

Hint: 

* Compare the given line to the equation of a straight line $$L$$ passing through the point $$(x_{1}, y_{1}, z_{1})$$ , to find the point through which the line passes and its directional ratio's.

* For a line to be parallel to a plane, then the line must be perpendicular to the normal of the plane.

Given:

Line: $$\dfrac{x-2}{3}=\dfrac{y-3}{4}=\dfrac{z-4}{5} \longrightarrow (i)$$ .

Step 1: Find the Direction Ratio's (DR's) of the given line $$L$$ .

We know, the equation of a straight line $$L$$ passing through the point $$(x_{1}, y_{1}, z_{1})$$ is:

$$L: \dfrac{x-x_{1}}{l}=\dfrac{y-y_{1}}{m}=\dfrac{z-z_{1}}{n} \longrightarrow (ii)$$ .

Comparing $$(i)$$ and $$(ii)$$ ,

we get, $$l=3,m=4,n=5 \longrightarrow (iii)$$ .

Step 2: Find the point through which the line $$L$$ passes through.

Now, comparing $$(i)$$ and $$(ii)$$ ,

we get,  $$x_{1}=2, y_{1}=3, z_{1}=4 \longrightarrow (iv)$$  .

$$\therefore$$ , The given line $$L$$ passes through the point  $$(2, 3, 4)$$ .

Hence, option $$D$$ is wrong.

Step 3: Check whether the given line $$L$$ lies in the plane $$3x+2y+6z-12=0$$ .

We know, the straight line $$L$$ lies in the plane if its point satisfies the equation of the plane.

Given, the plane is  $$3x+2y+6z-12=0 \longrightarrow (v)$$ .

Substitute $$(iv)$$ in $$(v)$$ , we get,

          $$3(2)+2(3)+6(4)-12=0$$  

$$\implies$$ $$6+6+24-12=0$$  

$$\implies$$ $$24\ne0$$ .

$$\therefore$$  The given line $$L$$ does not lie in the plane $$3x+2y+6z-12=0$$ .

Hence, option $$A$$ is wrong.

Step 4: Check whether the given line $$L$$ is perpendicular to the plane $$4x+7y+6z=0$$ .

We know, for a line to be perpendicular to a plane, then the line must be parallel to the normal of the plane.

Given, the plane  $$4x+7y+6z=0 \longrightarrow (vi)$$ .

Here, direction ratio's of the plane are $$l_{1}=4,m_{1}=7, n_{1}=6 \longrightarrow (vii)$$ .

Also, if  $$l_{1},m_{1}, n_{1}$$ and $$l,m,n$$  are the direction ratios of normal to plane and line $$L$$ ,

then  $$\dfrac{l_{1}}{l}=\dfrac{m_{1}}{m}=\dfrac{n_{1}}{n} \longrightarrow (viii)$$ if they are parallel.

Substitute $$(iii)$$ and $$(vii)$$ in $$(vi)$$ , we get,

                 $$\dfrac{4}{3}=\dfrac{7}{4}=\dfrac{6}{5}$$

But since  $$\dfrac{4}{3}\ne\dfrac{7}{4}\ne\dfrac{6}{5}$$ , 

$$\therefore$$  The given line $$L$$  is not perpendicular to the plane $$4x+7y+6z=0$$ .

Hence, option $$C$$ is wrong.

Step 5: Check whether the given line $$L$$ is parallel to the plane $$2x+y-2z=11$$ .

We know, for a line to be parallel to a plane, then the line must be perpendicular to the normal of the plane.

Given, the plane  $$2x+y-2z=11 \longrightarrow (ix)$$ .

Here, direction ratio's of the plane are $$l_{1}=2,m_{1}=1, n_{1}=-2 \longrightarrow (x)$$ .

Also, if  $$l_{1},m_{1}, n_{1}$$ and $$l,m,n$$  are the direction ratios of normal to plane and line $$L$$ ,

then  $$l_{1}l+m_{1}m+n_{1}n=0 \longrightarrow (xi)$$ if they are perpendicular.

Substitute $$(iii)$$ and $$(x)$$ in $$(xi)$$ , we get,

           $$(2)(3)+(1)(4)+(-2)(5)=0$$

$$\implies$$   $$6+4-10=0$$

$$\implies$$   $$0=0$$ .

$$\therefore$$  The given line $$L$$  is parallel to the plane $$2x+y-2z=11$$ .

Hence, option $$B$$ is correct.

Final Step:

The given line $$L$$  is parallel to the plane $$2x+y-2z=11$$ .

Hence, option $$B$$ is correct.