Distance between parallel planes MCQ Quiz in தமிழ் - Objective Question with Answer for Distance between parallel planes - இலவச PDF ஐப் பதிவிறக்கவும்

Concept

The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by the formula:

$\text{Distance}=\frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$

Calculation

$3=\frac{|a-1|}{\sqrt{2^2+1^2+1^2}}$

⇒ $3=\frac{|a-1|}{\sqrt{4+1+1}}$

⇒ $3=\frac{|a-1|}{\sqrt{6}}$

⇒ $3\sqrt{6}=|a-1|$

⇒ $a-1=3\sqrt{6}$ or $a-1=-3\sqrt{6}$.

Case 1: $a-1=3\sqrt{6}$

$a=1+3\sqrt{6}$

Case 2: $a-1=-3\sqrt{6}$

$a=1-3\sqrt{6}$

The product of all possible values of a is:

$(1+3\sqrt{6})(1-3\sqrt{6})=1^2-(3\sqrt{6}{)}^2=1-9(6)=1-54=-53$

Hence option 4 is correct

Concept Used:

The distance between two parallel planes $ax+by+cz=d_1$ and $ax+by+cz=d_2$ is given by $\frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$.

Calculation

Given:

Plane 1: $2x+3y+4z=4$

Plane 2: $4x+6y+8z=12$

We can rewrite the second equation by dividing by 2: $2x+3y+4z=6$

Now the equations are in the form $ax+by+cz=d$.

Here, $a=2$, $b=3$, $c=4$, $d_1=4$, and $d_2=6$.

$\text{Distance}=\frac{|6-4|}{\sqrt{2^2+3^2+4^2}}$

$\text{Distance}=\frac{2}{\sqrt{4+9+16}}$

$\text{Distance}=\frac{2}{\sqrt{29}}$

Therefore, the distance between the two planes is $\frac{2}{\sqrt{29}}$ units.

Hence option 3 is correct