Problem 56 - A Complete Resource Book in Mathematics for JEE Main

Question

A non-zero vector \(a\) is parallel to the line of intersection of the plane determined by the vectors \(i, i+j\) and the plane determined by the vectors \(i-j, i+k\). The angle between \(a\) and the vector \(i-2 j+2 k\) is (A) \(\frac{\pi}{3}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{3 \pi}{4}\) (D) none of these

Step-by-Step Solution

Answer

(D) none of these

Step 1: Identify Normal Vectors to Planes

The first plane is determined by the vectors \(i\) and \(i + j\). Vectors in the plane can be written as linear combinations of these two vectors. Similarly, the second plane is determined by the vectors \(i - j\) and \(i + k\). We find a normal vector to each plane by calculating the cross product of the two vectors spanning the plane.

Step 2: Compute Normal Vector for First Plane

For the plane determined by \(i\) and \(i + j\), we find the normal vector using the cross product: \(i \times (i + j)\). The cross product is only dependent on the results of \(i \times j\), as \(i \times i = 0\).\[i \times (i + j) = i \times i + i \times j = 0 + k = k\]Thus, the normal vector for the first plane is \(k\).

Step 3: Compute Normal Vector for Second Plane

For the second plane determined by \(i - j\) and \(i + k\), we calculate the cross product: \((i - j) \times (i + k)\).\[(i - j) \times (i + k) = i \times i + i \times k - j \times i - j \times k\]\[= 0 + j + k - (-k) = j + 2k\]Thus, the normal vector for the second plane is \(j + 2k\).

Step 4: Find Direction Vector of Line of Intersection

The line of intersection of the two planes is parallel to the cross product of their normal vectors. Calculate:\(k \times (j + 2k)\).\[= k \times j + 2k \times k\]\[= -i + 0 = -i\]Thus, the line of intersection is parallel to the vector \(-i\).

Step 5: Determine Vector \(a\)

Vector \(a\) must be a scalar multiple of \(-i\), since it needs to be parallel to the line of intersection. We can represent \(a\) as \(-i\).

Step 6: Calculate Angle Between \(a\) and \(i - 2j + 2k\)

The angle \(\theta\) between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by:\[\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\]Substitute \(\mathbf{u} = -i\) and \(\mathbf{v} = i - 2j + 2k\):Calculate the dot product:\(-i \cdot (i - 2j + 2k) = -1\)Calculate magnitudes:\(\|-i\| = 1\), \(\|i - 2j + 2k\| = \sqrt{1^2 + (-2)^2 + 2^2} = 3\)Substitute into formula:\[\cos \theta = \frac{-1}{1 \cdot 3} = -\frac{1}{3}\]Thus, \(\theta\) is \(\cos^{-1}(-\frac{1}{3})\), which is not related to any typical angles provided.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector algebra, used primarily in 3D space to determine a new vector that is orthogonal to two given vectors. For two vectors \(\mathbf{u} = a_1\mathbf{i} + b_1\mathbf{j} + c_1\mathbf{k}\) and \(\mathbf{v} = a_2\mathbf{i} + b_2\mathbf{j} + c_2\mathbf{k}\), the cross product \(\mathbf{u} \times \mathbf{v}\) is calculated as follows:

• Calculate the determinant of a 3x3 matrix:

\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix}\]

• The resulting vector is \([(b_1c_2 - c_1b_2)\mathbf{i} - (a_1c_2 - c_1a_2)\mathbf{j} + (a_1b_2 - b_1a_2)\mathbf{k} ]\).

In our exercise, we use the cross product to find normal vectors to planes, which helps us determine the direction of the line of intersection. It's essential in physics and engineering to reaffirm the perpendicularity and calculate torque.

Angle Between Vectors

The angle between two vectors is an important measurement that tells us about the vectors' directions relative to each other. To find this angle, we use the formula:

• \(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\)

Here, \(\mathbf{u} \cdot \mathbf{v}\) represents the dot product of the vectors, and \(\|\mathbf{u}\|\), \(\|\mathbf{v}\|\) are the magnitudes of the vectors.
The dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as \(a_1a_2 + b_1b_2 + c_1c_2\). The magnitude \(\|\mathbf{u}\|\) is given by \(\sqrt{a_1^2 + b_1^2 + c_1^2}\).
In the exercise, this formula helps us determine the angle between vector \(a\) and vector \(i - 2j + 2k\). This is especially useful in various applications such as navigation, computer graphics, and physics.

Plane Intersection

When two planes intersect, their line of intersection can be determined by the cross product of their normal vectors. The basic idea is that the direction of the line created by two intersecting planes is perpendicular to these normal vectors.
In the given exercise, two planes are determined by sets of vectors, and we first find the normal vectors to each:

• Calculate the cross product of the vectors that span each plane to find their normal vectors.
• For example, the planes spanned by \(i\), \(i+j\) and \(i-j\), \(i+k\) have normal vectors \(k\) and \(j+2k\), respectively.

The line of intersection we seek is parallel to the cross product of these normal vectors, i.e., \(k \times (j + 2k) = -i\), which guides us in determining vector \(a\). Understanding intersections is pivotal in graphics programming, modeling, and designing complex systems.