Distance between parallel planes MCQ Quiz in मल्याळम - Objective Question with Answer for Distance between parallel planes - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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