Problem 51 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The distance of the point \((1,-2,3)\) from the plane \(x-y+z=5\) measured parallel to the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\) is: \(\quad\) [NA Sep.04, 2020 (II)] (a) \(\frac{7}{5}\) (b) 1 (c) \(\frac{1}{7}\) (d) 7

Step-by-Step Solution

Answer

The distance is 1.

Step 1: Identify the Direction Ratios of the Line

The problem states that distance is measured parallel to the line \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \). This implies that the direction ratios of the line are \((2, 3, -6)\).

Step 2: Compute the Perpendicular Distance from Point to Plane

The perpendicular distance from a point \( (x_1, y_1, z_1) \) to a plane \( ax + by + cz = d \) is given by:\[\frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}.\]For the plane \( x - y + z = 5 \) and point \((1, -2, 3)\), substitute:\( a = 1, b = -1, c = 1, d = 5\), \( x_1 = 1, y_1 = -2, z_1 = 3 \).Calculate:\[\text{Numerator} = |1(1) -1(-2) + 1(3) - 5| = |1 + 2 + 3 - 5| = |1| = 1.\]\[\text{Denominator} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}.\]Perpendicular distance:\[\frac{1}{\sqrt{3}}.\]

Step 3: Find Projection of Perpendicular on Line

The projection of a vector along another vector is given by:\[\frac{\text{dot product of vectors}}{\text{magnitude of direction vector}}.\]Direction vector is \((2, 3, -6)\) with magnitude \(\sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49} = 7\).The dot product of \( (1/\sqrt{3})(1, -1, 1) \) and \((2, 3, -6)\):\[\frac{1}{\sqrt{3}}(2 - 3 - 6) = \frac{-7}{\sqrt{3}}.\]Projection length on line:\[\left|\frac{-7}{7}\right| = 1.\]

Step 4: Conclude the Calculated Distance

Now, the magnitude of the distance measured parallel to the line is equal to the magnitude of the projection calculated in the previous step, which is 1.

Key Concepts

Direction Ratios

Direction ratios are crucial in understanding lines in three-dimensional geometry. Consider direction ratios as the components that define the direction of a line.
In the given exercise, the line is described by the equation \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \). These fractions tell us how each coordinate component (\(x\), \(y\), \(z\)) changes in relation to each other.

• The direction ratios for this line are \((2, 3, -6)\).
• This means if \(x\) moves by 2 units, \(y\) changes by 3 units, and \(z\) decreases by 6 units.

Understanding these ratios helps us conceptualize and work with lines in space, allowing us to measure distances and angles effectively.

Perpendicular Distance Formula

The perpendicular distance from a point to a plane helps us understand how far a point is vertically from the plane's surface.
We use a specific formula for this distance:\[\frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\]This is derived from the equation of the plane \(ax + by + cz = d\). Each component of this formula represents:

• \(a, b, c\): Coefficients from the plane's equation.
• \((x_1, y_1, z_1)\): Coordinates of the point.
• The numerator involves plugging the point's coordinates into the plane equation and taking the absolute value.
• The denominator is the magnitude of the normal vector to the plane.

In this exercise, for the plane \(x - y + z = 5\) and point \((1, -2, 3)\), the distance is found to be \(\frac{1}{\sqrt{3}}\). This tells us the point is \(\frac{1}{\sqrt{3}}\) units pulled out, perpendicular to the plane.

Projection of a Vector

Vector projection is a way of decomposing one vector along the direction of another.This concept helps us determine how much of one vector lies along another, which is particularly useful in calculating distances parallel to lines.
We use the formula for projection:\[\frac{\text{dot product of vectors}}{\text{magnitude of direction vector}}\]In our exercise, we compute:

• The dot product of the vector \(\frac{1}{\sqrt{3}}(1, -1, 1)\) and direction vector \((2, 3, -6)\).
• The dot product gives \(\frac{-7}{\sqrt{3}}\).

Then divide this result by the magnitude of the direction vector, which is 7, to get the projection length along the line.The final answer is 1, indicating the point's location parallel to the line is 1 unit from the plane.