Key Concepts and Formulas

• Projection of a Line Segment onto a Plane: The projection of a line segment $P_1P_2$ of length $L$ onto a plane is the line segment $P_1'P_2'$, where $P_1'$ and $P_2'$ are the feet of the perpendiculars from $P_1$ and $P_2$ to the plane.
• Geometric Formula for Projection Length: If $\theta$ is the acute angle between the line segment and the plane, the length of the projection is $L \cos \theta$.
• Vector Formula for Projection Length: It is often more convenient to use the angle with the plane's normal. If $\vec{v}$ is the vector representing the line segment and $\vec{n}$ is the normal vector to the plane, let $\phi$ be the acute angle between $\vec{v}$ and $\vec{n}$. Then, $\theta = 90^\circ - \phi$. Therefore, the length of the projection is $L \cos(90^\circ - \phi) = L \sin \phi$.
• Angle Between Two Vectors (Dot Product): For two vectors $\vec{a}$ and $\vec{b}$, the cosine of the angle $\phi$ between them is given by: $\cos \phi = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$ We use the absolute value of the dot product to ensure $\phi$ is the acute angle.
• Pythagorean Identity: $\sin^2 \phi + \cos^2 \phi = 1$.
• Equation of a Line Perpendicular to a Plane: A line passing through a point $(x_0, y_0, z_0)$ and perpendicular to a plane $Ax+By+Cz=D$ has parametric equations: $x = x_0 + At$, $y = y_0 + Bt$, $z = z_0 + Ct$.

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Step-by-Step Solution

We will solve this problem using two methods. The first method, utilizing the angle between the line segment and the plane's normal vector, is generally more efficient for JEE problems.

Method 1: Using the Angle with the Normal Vector (Recommended Approach)

This method directly applies the vector formula for projection length.

Step 1: Define the line segment vector and find its length. We are given two points $P_1(5, -1, 4)$ and $P_2(4, -1, 3)$. To represent the line segment $P_1P_2$ as a vector, we find the displacement vector from $P_1$ to $P_2$: $\vec{v} = \vec{P_1P_2} = P_2 - P_1 = (4-5, -1-(-1), 3-4) = (-1, 0, -1)$ The length of the line segment $L$ is the magnitude of this vector: $L = |\vec{v}| = \sqrt{(-1)^2 + (0)^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$ Explanation: The vector $\vec{v}$ defines the direction and magnitude of the line segment. Its magnitude $L$ is the actual length of the segment in 3D space, which is a key component in our projection formula.

Step 2: Identify the plane's normal vector and its magnitude. The equation of the plane is $x + y + z = 7$. For a plane in the form $Ax + By + Cz = D$, the normal vector is $\vec{n} = (A, B, C)$. Thus, the normal vector to this plane is: $\vec{n} = (1, 1, 1)$ The magnitude of the normal vector is: $|\vec{n}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$ Explanation: The normal vector is perpendicular to the plane. The angle between the line segment and this normal vector is critical because it's complementary to the angle between the line segment and the plane itself, simplifying the projection calculation.

Step 3: Calculate the cosine of the acute angle ($\phi$) between $\vec{v}$ and $\vec{n}$. Using the dot product formula for the angle between two vectors: $\cos \phi = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$ First, calculate the dot product: $\vec{v} \cdot \vec{n} = (-1)(1) + (0)(1) + (-1)(1) = -1 + 0 - 1 = -2$ Now, substitute the values into the formula for $\cos \phi$: $\cos \phi = \frac{|-2|}{\sqrt{2} \cdot \sqrt{3}} = \frac{2}{\sqrt{6}}$ Explanation: The dot product quantifies the alignment of two vectors. We take the absolute value of the dot product to ensure that $\phi$ is the acute angle, which is consistent with the geometric definition of the angle between a line and a plane.

Step 4: Calculate $\sin \phi$. We need $\sin \phi$ for the projection formula. We can find it using the Pythagorean identity $\sin^2 \phi + \cos^2 \phi = 1$: $\sin^2 \phi = 1 - \cos^2 \phi = 1 - \left(\frac{2}{\sqrt{6}}\right)^2 = 1 - \frac{4}{6} = 1 - \frac{2}{3} = \frac{1}{3}$ Since $\phi$ is an acute angle (as we ensured by using the absolute value in $\cos \phi$), $\sin \phi$ must be positive: $\sin \phi = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$ Explanation: This step converts the cosine value (derived from the dot product) into the sine value, which is directly used in our chosen projection formula ($L \sin \phi$).

Step 5: Calculate the length of the projection. Using the formula derived from key concepts: $\text{Length of projection} = L \sin \phi$ Substitute the values of $L = \sqrt{2}$ and $\sin \phi = \frac{1}{\sqrt{3}}$: $\text{Length of projection} = \sqrt{2} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$

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Method 2: Finding the Feet of Perpendiculars (Alternative Approach)

This method involves finding the coordinates of the projections of the endpoints of the line segment onto the plane, and then calculating the distance between these projected points.

Step 1: Find the foot of the perpendicular from $P_1(5, -1, 4)$ to the plane $x+y+z=7$. The direction vector of the line perpendicular to the plane is the normal vector of the plane, $\vec{n} = (1, 1, 1)$. The parametric equations of the line passing through $P_1$ and perpendicular to the plane are: $L_1: x = 5+t, \quad y = -1+t, \quad z = 4+t$ To find the foot of the perpendicular $P_1'$, substitute these parametric equations into the plane equation: $(5+t) + (-1+t) + (4+t) = 7$ $8 + 3t = 7 \implies 3t = -1 \implies t = -\frac{1}{3}$ Substitute $t = -1/3$ back into the parametric equations to find the coordinates of $P_1'$: $P_1' = \left(5 - \frac{1}{3}, -1 - \frac{1}{3}, 4 - \frac{1}{3}\right) = \left(\frac{15-1}{3}, \frac{-3-1}{3}, \frac{12-1}{3}\right) = \left(\frac{14}{3}, -\frac{4}{3}, \frac{11}{3}\right)$

Step 2: Find the foot of the perpendicular from $P_2(4, -1, 3)$ to the plane $x+y+z=7$. Similarly, for $P_2$, the parametric equations of the line perpendicular to the plane are: $L_2: x = 4+s, \quad y = -1+s, \quad z = 3+s$ Substitute these into the plane equation: $(4+s) + (-1+s) + (3+s) = 7$ $6 + 3s = 7 \implies 3s = 1 \implies s = \frac{1}{3}$ Substitute $s = 1/3$ back into the parametric equations to find the coordinates of $P_2'$: $P_2' = \left(4 + \frac{1}{3}, -1 + \frac{1}{3}, 3 + \frac{1}{3}\right) = \left(\frac{12+1}{3}, \frac{-3+1}{3}, \frac{9+1}{3}\right) = \left(\frac{13}{3}, -\frac{2}{3}, \frac{10}{3}\right)$

Step 3: Calculate the distance between $P_1'$ and $P_2'$. The length of the projection is the distance between the two projected points $P_1'$ and $P_2'$: $\text{Length} = \sqrt{\left(\frac{14}{3} - \frac{13}{3}\right)^2 + \left(-\frac{4}{3} - \left(-\frac{2}{3}\right)\right)^2 + \left(\frac{11}{3} - \frac{10}{3}\right)^2}$ $= \sqrt{\left(\frac{1}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2}$ $= \sqrt{\frac{1}{9} + \frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$ Both methods yield the same result.

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Common Mistakes & Tips

• Confusing Angles: A common mistake is using the angle between the line segment and the plane's normal vector directly as $\theta$ in $L \cos \theta$. Remember, the angle with the plane is complementary to the angle with the normal vector. So, if using the normal vector, the formula is $L \sin \phi$.
• Absolute Value in Dot Product: Always use the absolute value of the dot product (i.e., $|\vec{v} \cdot \vec{n}|$) when calculating $\cos \phi$ to ensure you find the acute angle between the line segment and the normal vector. This prevents issues if the initial direction of the vector happens to point "away" from the normal.
• Calculation Errors: Pay close attention to arithmetic, especially when dealing with fractions in the second method. A small error can lead to a completely different answer.
• Choosing the Right Method: For projection length problems, the vector method (Method 1) is generally quicker and less prone to calculation errors involving fractions, making it the preferred approach in time-sensitive exams like JEE.

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Summary

The problem asked for the length of the projection of a line segment onto a plane. We successfully solved this using the vector method, which involves finding the length of the line segment ($L$) and the acute angle ($\phi$) between the segment's direction vector and the plane's normal vector. The length of the projection is then given by $L \sin \phi$. This approach is efficient and less prone to errors compared to finding the feet of perpendiculars. Both methods confirmed the same result.

The final answer is $\boxed{\sqrt {{2 \over 3}}}$, which corresponds to option (A).