Problem 131 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The d.r. of normal to the plane through \((1,0,0),(0,1,0)\) which makes an angle \(\pi / 4\) with plane \(x+y=3\) are (a) \(1, \sqrt{2}, 1\) (b) \(1,1, \sqrt{2}\) (c) \(1,1,2\) (d) \(\sqrt{2}, 1,1\)

Step-by-Step Solution

Answer

The direction ratios are \(1, 1, \sqrt{2}\) (option b).

Step 1: Identify Normal to the Given Plane

The equation of a plane is given by \(x + y = 3\). The direction ratios of the normal to this plane are coefficients of \(x\) and \(y\), which are \(1\) and \(1\), respectively. Hence, the normal vector to the plane \(x + y = 3\) is \(\hat{n_1} = (1, 1, 0)\).

Step 2: Find the Normal to the Plane through Given Points

A plane through the points \((1,0,0)\) and \((0,1,0)\) has a direction vector which can be obtained by subtracting these points, giving \((0-1, 1-0, 0-0) = (-1, 1, 0)\). So, this vector, \(\mathbf{v} = (-1,1,0)\), lies on the plane.

Step 3: Compute the Arbitrary Normal Vector to the New Plane

Assume a direction vector \(\mathbf{d} = (a, b, c)\) which is perpendicular to the direction vector \((-1, 1, 0)\) lying on the plane. Since \(\mathbf{d}\) is perpendicular, it must satisfy: \((-1, 1, 0) \cdot (a, b, c) = 0\) leading to \(-a + b = 0\) or \(b = a\).

Step 4: Use Angle Condition to Find Possible Directions

The normal vector \(\mathbf{d} = (a, b, c)\) makes an angle \(\frac{\pi}{4}\) with \(\hat{n_1} = (1, 1, 0)\). Using the dot product formula, \[ \cos(\theta) = \frac{\mathbf{d} \cdot \hat{n_1}}{|\mathbf{d}| \cdot |\hat{n_1}|} \] and \(\theta = \frac{\pi}{4}\) the given implies \[ \frac{a+b}{\sqrt{a^2+b^2+c^2} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}.\]

Step 5: Solve the Cosine Condition

The equation from the angle condition becomes \[ (a+b) = \frac{1}{\sqrt{2}} \cdot \sqrt{2(a^2+b^2+c^2)}, \]which simplifies using \(b = a\), \[ 2a = \sqrt{a^2+a^2+c^2}.\]Squaring both sides gives \(4a^2 = 2a^2 + c^2\) leading to \(2a^2 = c^2\).

Step 6: Solution Determination and Selection of Desired Ratios

Solve for \(c\) in terms of \(a\), giving \(c = \sqrt{2}a\).The direction ratios satisfy \((a, a, \sqrt{2}a)\), which can be normalized and compared with options. Choose option (b): \(1, 1, \sqrt{2}\).

Key Concepts

Direction Ratios

Direction ratios play a crucial role in representing the orientations and directions of vectors in geometry. They are essentially the coefficients of a vector, indicating its direction. For example, in the context of a plane, if a normal vector has coefficients

• x,
• y,
• z,

this means that the vector formed by a x y. indicates the direction normal to the plane. When given the equation of a plane, such as

• x + y = 3,

the direction ratios for the normal vector are derived from the coefficients of x and y, which in this case are

• 1 and
• 1,

forming the vector (1, 1, 0).

Normal to Plane

The concept of the normal to a plane is fundamental when working with geometrical shapes and vectors, especially in 3D space. A normal vector is perpendicular to the surface of the plane, much like a stick standing upright from a flat surface. Determining the normal vector gives insight into how the plane is oriented.

• To find the normal to a plane using points, subtracting one point from another can help create a vector that lies on the plane itself.
• One can then use vector cross-products to determine a vector perpendicular to both the resultant vector and the plane.

In some simpler cases, simply identifying the coefficients of each variable in the plane's equation can instantly give us these direction ratios, indicated by the normal vector. For example, in the problem given, the normal vector to the plane defined by

• x + y = 3

is calculated directly as (1, 1, 0). This vector is perpendicular to any vector lying on the plane such as

• (-1, 1, 0),

as calculated from the points provided.

Dot Product

The dot product, also known as the scalar product, is significant in understanding angles between vectors. It represents how much of one vector goes in the direction of another, essentially giving a measure of their projection onto one another. When calculating a dot product between two vectors, the formula is:\[\text{Dot Product} = \mathbf{a} \cdot \mathbf{b} = x_1x_2 + y_1y_2 + z_1z_2\]This multiplication ties into the calculation of the angle as it correlates with the magnitudes of the vectors and the cosine of the angle between them, which is:\[\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \: |\mathbf{b}| \: \cos(\theta)\] When working on the plane and vector problems, using the dot product allows us not only to determine if vectors are perpendicular (i.e., when\[\mathbf{a} \cdot \mathbf{b} = 0\])but also find the precise angle they form.

Angle Between Vectors

The angle between vectors is a crucial concept, especially in geometrical problems involving planes and angles. It helps determine the spatial orientation and can further assist in resolving whether planes or lines are parallel, perpendicular, or at another specific angular alignment.

• To compute the angle between vectors, the dot product formula ties in perfectly: \[\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \: |\mathbf{b}|}\]This formula shows the trigonometric function
• \(\cos(\theta)\), indicating that simplifying this equation can directly solve for \(\theta\).

In practical terms, recognizing that they have plotted an angle of
\(\pi/4\) radians imparts specific relationships between vectors—for instance, vectors making an angle of \(\pi/4\) often indicate complementary angles within geometric transformations.