Problem 40 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(|\vec{a}|=2,|\vec{b}|=3\) and \(|2 \vec{a}-\vec{b}|=5\), then \(|2 \vec{a}+\vec{b}|\) equals: [Online April 9, 2014] (a) 17 (b) 7 (c) 5 (d) 1

Step-by-Step Solution

Answer

Option (c) 5 is correct.

Step 1: Write vector magnitude formulas

Given the magnitudes of \(\vec{a}\) and \(\vec{b}\), we need to use the formula for the magnitude of vectors. The magnitude formula for any vector \(\vec{v}\) is \(|\vec{v}| = \sqrt{\vec{v} \cdot \vec{v}}\).

Step 2: Express given information as equations

We have \(|2\vec{a} - \vec{b}| = 5\) which means:\[|2\vec{a} - \vec{b}| = \sqrt{(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b})} = 5\]This can be expanded using the dot product as:\[4(\vec{a}\cdot\vec{a}) - 4(\vec{a}\cdot\vec{b}) + (\vec{b}\cdot\vec{b}) = 25\]

Step 3: Substitute known magnitudes

Substitute the given magnitudes of \(\vec{a}\) and \(\vec{b}\) into the equation:\[4(2^2) - 4(\vec{a}\cdot\vec{b}) + 3^2 = 25\] This simplifies to:\[16 - 4(\vec{a}\cdot\vec{b}) + 9 = 25\]

Step 4: Solve for dot product \(\vec{a} \cdot \vec{b}\)

Combine like terms and solve for \(\vec{a}\cdot\vec{b}\):\[25 - 4(\vec{a}\cdot\vec{b}) = 25\] \[-4(\vec{a}\cdot\vec{b}) = 0\] So, \(\vec{a}\cdot\vec{b} = 0\).

Step 5: Find \(|2\vec{a} + \vec{b}|\)

Now we need \(|2\vec{a} + \vec{b}|\):\[|2\vec{a} + \vec{b}| = \sqrt{(2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b})}\] Substitute known values:\[4(\vec{a}\cdot\vec{a}) + 4(\vec{a}\cdot\vec{b}) + (\vec{b}\cdot\vec{b})\]Since \(\vec{a} \cdot \vec{b} = 0\), the expression simplifies to:\[16 + 0 + 9 = 25\]\[\sqrt{25} = 5\]

Step 6: Conclusion

The magnitude \(|2\vec{a} + \vec{b}|\) is \(5\). Hence, the correct option is (c) 5.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It helps to calculate the magnitude of a vector or the angle between two vectors. The dot product of two vectors \(\vec{u}\) and \(\vec{v}\) is defined as:\[\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta\]where \(\theta\) is the angle between the two vectors.
The dot product yields a scalar quantity, which means it has only magnitude and no direction. It is often used to find orthogonal vectors, which are vectors with a dot product of zero.

• When two vectors are perpendicular, their dot product is zero: \(\vec{a} \cdot \vec{b} = 0\).
• Dot products are commutative, so \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\).
• They are also distributive, for example \((\vec{a} + \vec{c}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{b}\).

Understanding how to calculate and use the dot product is key in solving vector problems and analyzing vector relationships.

Vector Algebra

Vector algebra involves operations that include vectors, which are mathematical entities with both magnitude and direction. In vector mathematics, we employ operations such as addition, subtraction, and scalar multiplication. These operations help us solve problems related to physics, engineering, and more.
For addition and subtraction, we use the component-wise method. Two vectors \(\vec{a} = (a_1, a_2)\) and \(\vec{b} = (b_1, b_2)\) are added as follows:

• Add corresponding components: \(\vec{a} + \vec{b} = (a_1 + b_1, a_2 + b_2)\).
• For subtraction: \(\vec{a} - \vec{b} = (a_1 - b_1, a_2 - b_2)\).

Scalar multiplication is similar: if a vector \(\vec{a} = (a_1, a_2)\) is multiplied by a scalar \(k\), it becomes \(k\vec{a} = (ka_1, ka_2)\).
Vector algebra is crucial for understanding vector transformations and can be used to describe physical properties like force and velocity in physics. By understanding these operations, you can analyze the dynamics of vector quantities in real-world applications.

Magnitude of Vectors

The magnitude of a vector represents its length or size and is one of the most important properties of a vector in mathematics and physics. The formula to find a vector's magnitude \(|\vec{v}|\) is given by the square root of the dot product of the vector with itself:
\[|\vec{v}| = \sqrt{\vec{v} \cdot \vec{v}}\]For example, if a vector \(\vec{v} = (x, y)\), its magnitude is calculated as:
\[|\vec{v}| = \sqrt{x^2 + y^2}\]
Understanding vector magnitude is essential for applications in physics and engineering, such as calculating distances and measuring velocities. It gives you a sense of how large the vector is, irrespective of its direction.

• The magnitude of a vector is always a non-negative number.
• Magnitude is used in normalizing vectors, which creates a unit vector with a magnitude of 1.
• It is also pivotal in problems requiring the calculation of physical lengths and distances.

Mastering the concept of magnitudes will enhance your ability to work with vectors in various mathematical and physical contexts.