Problem 49 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The non-zero vectors are \(\vec{a}, \vec{b}\) and \(c\) are related by \(\vec{a}=8 \vec{b}\) and \(\vec{c}=-7 \bar{b}\). Then the angle between \(\vec{a}\) and \(\vec{c}\) is [2008] (a) 0 (b) \(\frac{\pi}{4}\) (c) \(\frac{\pi}{2}\) (d) \(\pi\)

Step-by-Step Solution

Answer

The angle between \(\vec{a}\) and \(\vec{c}\) is \(\pi\) radians, or 180 degrees.

Step 1: Understanding Vector Relations

We are given that \( \vec{a} = 8 \vec{b} \) and \( \vec{c} = -7 \vec{b} \). This means that \( \vec{a} \) is a scaled version of \( \vec{b} \) by a factor of 8, and \( \vec{c} \) is a scaled and direction-reversed version of \( \vec{b} \) by a factor of 7.

Step 2: Express \( \vec{a} \) and \( \vec{c} \) in Terms of \( \vec{b} \)

Both vectors \( \vec{a} \) and \( \vec{c} \) are collinear with \( \vec{b} \). Specifically, \( \vec{a} = 8 \vec{b} \) and \( \vec{c} = -7 \vec{b} \).

Step 3: Calculate the Dot Product

The dot product between two vectors \( \vec{a} \) and \( \vec{c} \) is given by: \( \vec{a} \cdot \vec{c} = 8 \vec{b} \cdot (-7 \vec{b}) = -56 (\vec{b} \cdot \vec{b}) \).

Step 4: Use the Dot Product to Find the Angle

The formula for the dot product in terms of the angle \( \theta \) between two vectors is given by \( \vec{a} \cdot \vec{c} = ||\vec{a}|| ||\vec{c}|| \cos \theta \). Since \( \vec{a} \) and \( \vec{c} \) are collinear and also have opposite directions (because \( \vec{c} = -7 \vec{b} \)), the angle between them is \( \pi \) radians, or 180 degrees.

Step 5: Conclude with the Angle

Given that two vectors are in opposite directions, the angle between them is \( \pi \). Thus, the correct answer is option (d) \( \pi \).

Key Concepts

Angle Between Vectors

The angle between two vectors is an important concept in vector mathematics. It is described by the formula for the dot product in terms of the angle \( \theta \), which is \( \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos \theta \). This formula means the dot product is the product of the magnitudes of the two vectors and the cosine of the angle between them.
To find the angle, rearrange the formula as follows:

• Calculate the dot product of the vectors.
• Divide by the product of the magnitudes of the vectors.
• Use the arccos function to find the angle \( \theta \).

When vectors are collinear but point in opposite directions, like in this example with \( \vec{a} \) and \( \vec{c} \), the angle between them is \( \pi \) radians (180 degrees). This is because the dot product will be negative, leading to a cosine of \(-1\).

Collinear Vectors

Vectors are collinear when they lie on the same line or are parallel to each other. This means one vector is a scalar multiple of the other. Here, \( \vec{a} \) and \( \vec{c} \) are collinear with \( \vec{b} \) because both are scaled versions of \( \vec{b} \):\( \vec{a} = 8 \vec{b} \) and \( \vec{c} = -7 \vec{b} \).
This relation shows:

• \( \vec{a} \) extends in the same direction as \( \vec{b} \), but is 8 times its magnitude.
• \( \vec{c} \) extends in the opposite direction as \( \vec{b} \), but is 7 times its magnitude.

Collinear vectors are simpler to work with because their directions are either the same or directly opposite, making calculations straightforward.

Dot Product Calculation

The dot product is a measure of how much two vectors "point" in the same direction. It is calculated by multiplying the corresponding components of the two vectors and then summing those products. For example, the dot product formula is \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \) for 3D vectors.
In this problem, because both \( \vec{a} \) and \( \vec{c} \) are defined in terms of \( \vec{b} \), their dot product simplifies to \( \vec{a} \cdot \vec{c} = (8 \vec{b}) \cdot (-7 \vec{b}) = -56 (\vec{b} \cdot \vec{b}) \).
Here, the negative sign indicates the opposite directions of \( \vec{a} \) and \( \vec{c} \), and \( \vec{b} \cdot \vec{b} \) is simply the magnitude of \( \vec{b} \) squared.

Vector Magnitude

The magnitude of a vector, also known as its length or norm, is indicated by the symbols \( \|\vec{a}\| \). It quantifies the size of the vector without considering its direction.
For a vector \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), its magnitude is calculated using the formula:

• \( \|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

In the given situation, even though we didn't compute specific magnitudes, knowing their values helps when understanding the formula for the dot product, where \( \vec{a} \cdot \vec{c} = \|\vec{a}\| \|\vec{c}\| \cos \theta \).
Recognizing vector magnitudes allows you to better analyze the properties and relationships between vectors.

Scalar Multiplication of Vectors

Scalar multiplication involves multiplying a vector by a scalar (a real number), which scales the magnitude of the vector while preserving its direction (or reversing it if the scalar is negative).
For a vector \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), and a scalar \( k \), the scalar multiplication is:

• \( k \vec{v} = \langle kv_1, kv_2, kv_3 \rangle \)

In the exercise, \( \vec{a} = 8 \vec{b} \) and \( \vec{c} = -7 \vec{b} \) demonstrate scalar multiplication. This operation results in vectors that are longer or shorter than the original based on the scalar’s value. It's a fundamental concept in vector algebra, showing how vectors can be manipulated to form new vectors.