Question

A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the co-ordinate axes. The line meets the plane $2x+y+z=9$ at point Q. The length of the line segment PQ equals

• A. 3
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. 2

Hint:

Geometry Tip: When a line makes equal angles with coordinate axes, its direction ratios can immediately be taken as $(1, 1, 1)$. You don't need to carry the fractions $\frac{1}{\sqrt{3}}$ through your line equation calculations.

Answer 1

The Correct Option is C

Concept: 3D Geometry - Direction Cosines and Line-Plane Intersection.

Step 1: Determine the direction cosines of the line. The line makes equal angles with the coordinate axes, so its direction angles satisfy $\alpha = \beta = \gamma$. Consequently, the direction cosines are also equal: $l = m = n = \cos\alpha$. Using the fundamental identity $l^2 + m^2 + n^2 = 1$, we get $3\cos^2\alpha = 1 \implies \cos^2\alpha = \frac{1}{3}$. Since the direction cosines are given as positive, $\cos\alpha = \frac{1}{\sqrt{3}}$. Thus, the direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Step 2: Formulate the equation of the line. The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction cosines $(l, m, n)$ is $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$. Substitute point $P(2, -1, 2)$ and our derived direction cosines: $\frac{x-2}{\frac{1}{\sqrt{3}}} = \frac{y+1}{\frac{1}{\sqrt{3}}} = \frac{z-2}{\frac{1}{\sqrt{3}}}$. Multiplying the denominators by $\sqrt{3}$ yields equivalent direction ratios of $(1, 1, 1)$. The simplified line equation is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1}$.

Step 3: Define the coordinates of the general point Q. Set the line equation equal to a scalar constant $k$: $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$. Express $x, y, z$ in terms of $k$ to find the coordinates of any point on the line, which we will call Q: $Q \equiv (k+2, k-1, k+2)$.

Step 4: Find the intersection point with the plane. Point Q lies on the given plane $2x+y+z=9$. Substitute the coordinates of Q into the plane's equation: $2(k+2) + (k-1) + (k+2) = 9$. Expand and simplify: $2k + 4 + k - 1 + k + 2 = 9$.
Combine like terms: $4k + 5 = 9 \implies 4k = 4 \implies k = 1$. Substitute $k=1$ back into our point definition to get the exact coordinates: $Q \equiv (3, 0, 3)$.

Step 5: Calculate the length of the line segment PQ. Use the 3D distance formula to find the length between $P(2, -1, 2)$ and $Q(3, 0, 3)$: $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. $PQ = \sqrt{(3-2)^2 + (0-(-1))^2 + (3-2)^2}$. $PQ = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$. $$ \therefore \text{The length of the line segment PQ is } \sqrt{3}. $$