Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The position vectors of two points \(A\) and \(C\) are \(9 i-j+\) \(7 k\) and \(7 i-2 j+7 k\), respectively. The point of intersection of vectors \(A B=4 i-j+3 k\) and \(C D=2 i-j+2 k\) is \(P\). If vector \(P Q\) is perpendicular to \(A B\) and \(C D\) and \(P Q\) \(=15\) units, the position vector of \(Q\) is (A) \(6 i-9 j-9 k\) (B) \(-4 i+11 j+11 k\) (C) \(6 i+9 j-9 k\) (D) none of these

Step-by-Step Solution

Answer

The position vector of Q is not an option; it's 22i - 13j + 3k.

Step 1: Determine Vector AB and CD

The vectors for AB and CD are given: \( \mathbf{AB} = 4i - j + 3k \) and \( \mathbf{CD} = 2i - j + 2k \). Ensure these vectors are correctly identified.

Step 2: Find the Intersection Point P

The intersection point \( P \) is found by observing that points along \( AB \) and \( CD \) can be parameterized. If we set \( \mathbf{A} + t\mathbf{AB} = \mathbf{C} + s\mathbf{CD} \), equate terms:- For \( i \): \( 9 + 4t = 7 + 2s \)- For \( j \): \( -1 - t = -2 - s \)- For \( k \): \( 7 + 3t = 7 + 2s \)From \( j \), we find \( t = s - 1 \). Substitute into \( k \): \( 3(s-1) = 2s \), solving gives \( s = 3 \) and then \( t = 2 \). So, \( P = \mathbf{A} + 2\mathbf{AB} = 9i - j + 7k + 2(4i - j + 3k) = 17i - 3j + 13k \).

Step 3: Vector PQ Perpendicularity Condition

For \( \mathbf{PQ} \) to be perpendicular to both \( \mathbf{AB} \) and \( \mathbf{CD} \), its dot product with each must be zero. The cross product \( \mathbf{AB} \times \mathbf{CD} \) gives the direction perpendicular to both:\[ \mathbf{AB} \times \mathbf{CD} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & -1 & 3 \ 2 & -1 & 2 \end{array} \right| = \begin{vmatrix} -1 & 3 \ -1 & 2 \end{vmatrix} \mathbf{i} - \begin{vmatrix} 4 & 3 \ 2 & 2 \end{vmatrix} \mathbf{j} + \begin{vmatrix} 4 & -1 \ 2 & -1 \end{vmatrix} \mathbf{k} \]This simplifies to \( \mathbf{i} - 2\mathbf{j} - 2\mathbf{k} \).

Step 4: Determine Q Using Length of PQ

The magnitude of \( \mathbf{PQ} \) is 15, so \( \mathbf{PQ} = k(\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}) \), with \( k \) a scalar:\( |k(\mathbf{i} - 2\mathbf{j} - 2\mathbf{k})| = 15 \). The magnitude calculation gives:\[ |k| \sqrt{9} = 15 \rightarrow |k| = 5. \]Thus \( \mathbf{PQ} = 5(\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}) = 5\mathbf{i} - 10\mathbf{j} - 10\mathbf{k} \).

Step 5: Calculate Position Vector of Q

Given that \( \mathbf{Q} = \mathbf{P} + \mathbf{PQ} \) and \( P = 17i - 3j + 13k \), \[ \mathbf{Q} = (17i - 3j + 13k) + (5i - 10j - 10k) = 22i - 13j + 3k. \]This doesn’t match the options given, so the correct answer is (D) none of these.

Key Concepts

Position Vectors

Position vectors are a fundamental concept in vector algebra used to determine the location of points in space. They provide a way to precisely describe points relative to a reference point, usually the origin.

• In a three-dimensional coordinate system, position vectors are typically expressed in the form of \(ai + bj + ck\), where \(i, j, k\) are the unit vectors along the x, y, and z axes respectively.
• The coefficients \(a, b, c\) represent the coordinates of the point in 3D space.
• Position vectors are handy for performing operations such as calculating distances between points and evaluating other vectors originating from or terminating at given points.

In the provided exercise, the position vectors for points A and C are specified as \(9i - j + 7k\) and \(7i - 2j + 7k\) respectively. This implies that Point A is located 9 units along the x-axis, -1 unit along the y-axis, and 7 units along the z-axis from the origin. Similarly, Point C's respective position can be deduced.

Dot Product

The dot product, often known as the scalar product, is a fundamental operation in vector algebra used to determine if two vectors are orthogonal (perpendicular) to one another. The dot product results in a scalar quantity rather than a vector.

• The mathematical formula for the dot product of two vectors \(\mathbf{A} = ai + bj + ck\) and \(\mathbf{B} = di + ej + fk\) is: \(\mathbf{A} \cdot \mathbf{B} = ad + be + cf\).
• When the dot product is zero, it indicates that the vectors are perpendicular, meaning their angle of intersection is 90 degrees.
• This property is beneficial for determining relationships between directed lines and planes in space.

For vectors to be perpendicular, as required for finding PQ in the exercise, their dot products with other vectors must equal zero. This crucial condition was used to determine the direction of vector PQ which was required to be perpendicular to vectors AB and CD.

Cross Product

The cross product, or vector product, is an operation between two vectors in three-dimensional space. Unlike the dot product, it results in a vector that is perpendicular to both of the original vectors. It plays a significant role in physics and engineering, especially in problems involving rotational motion and forces.

• The cross product of two vectors \(\mathbf{A} = ai + bj + ck\) and \(\mathbf{B} = di + ej + fk\) is calculated using the determinant of a matrix of unit vectors and components: \[\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a & b & c \ d & e & f \end{vmatrix}\]
• This expands to \( (bf - ce)\mathbf{i} - (af-cd)\mathbf{j} + (ae-bd)\mathbf{k}\).
• The magnitude of the cross product vector represents the area of the parallelogram formed by the original two vectors.

In the exercise, the cross product of vectors AB and CD was utilized to calculate a perpendicular vector for PQ. The findings derived from this determined both the direction and subsequent calculations of the position vector of point Q.