Understanding Vector Perpendicularity and Magnitude

The question asks us to evaluate a vector dot product involving vectors \(\vec{a}\) and \(\vec{b}\) given two conditions: that the sum of the vectors, \(\vec{a} + \vec{b}\), is perpendicular to \(\vec{a}\), and that the magnitude of \(\vec{b}\) is twice the magnitude of \(\vec{a}\).

Let's break down the given information and what we need to find.

• Given Condition 1: \(\rm(\vec {a} + \vec{b})\) is perpendicular to \(\rm\vec {a}\).
• Given Condition 2: Magnitude of \(\rm\vec {b}\) is twice that of \(\rm\vec {a}\), which can be written as \(|\vec{b}| = 2|\vec{a}|\).
• To Find: The value of \(\rm(4\vec {a} + \vec{b})\cdot \vec{b}\).

Using the Perpendicularity Condition

When two vectors are perpendicular, their dot product is zero. So, the condition that \(\rm(\vec {a} + \vec{b})\) is perpendicular to \(\rm\vec {a}\) translates to:

\((\vec{a} + \vec{b}) \cdot \vec{a} = 0\)

Using the distributive property of the dot product, we can expand this:

\(\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} = 0\)

Recall that the dot product of a vector with itself is the square of its magnitude, i.e., \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\). Also, the dot product is commutative, so \(\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}\).

Substituting this into the equation:

\(|\vec{a}|^2 + \vec{a} \cdot \vec{b} = 0\)

From this, we can find a relationship between the dot product \(\vec{a} \cdot \vec{b}\) and the magnitude \(|\vec{a}|^2\):

\(\vec{a} \cdot \vec{b} = -|\vec{a}|^2\)

Using the Magnitude Condition

The second condition given is that the magnitude of \(\vec{b}\) is twice that of \(\vec{a}\):

\(|\vec{b}| = 2|\vec{a}|\)

Squaring both sides of this equation will be useful later, as dot products often involve squared magnitudes:

\(|\vec{b}|^2 = (2|\vec{a}|)^2\)

\(|\vec{b}|^2 = 4|\vec{a}|^2\)

Evaluating the Required Expression

We need to find the value of \(\rm(4\vec {a} + \vec{b})\cdot \vec{b}\). Let's expand this expression using the distributive property of the dot product:

\((4\vec{a} + \vec{b}) \cdot \vec{b} = (4\vec{a}) \cdot \vec{b} + \vec{b} \cdot \vec{b}\)

Using the property \((c\vec{a}) \cdot \vec{b} = c(\vec{a} \cdot \vec{b})\) where \(c\) is a scalar, and \(\vec{b} \cdot \vec{b} = |\vec{b}|^2\), we get:

\((4\vec{a} + \vec{b}) \cdot \vec{b} = 4(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\)

Now we can substitute the relationships we found from the given conditions:

Substitute \(\vec{a} \cdot \vec{b} = -|\vec{a}|^2\):

\((4\vec{a} + \vec{b}) \cdot \vec{b} = 4(-|\vec{a}|^2) + |\vec{b}|^2\)

\((4\vec{a} + \vec{b}) \cdot \vec{b} = -4|\vec{a}|^2 + |\vec{b}|^2\)

Substitute \(|\vec{b}|^2 = 4|\vec{a}|^2\):

\((4\vec{a} + \vec{b}) \cdot \vec{b} = -4|\vec{a}|^2 + 4|\vec{a}|^2\)

\((4\vec{a} + \vec{b}) \cdot \vec{b} = 0\)

The value of \(\rm(4\vec {a} + \vec{b})\cdot \vec{b}\) is 0.

Summary of Calculation Steps

Revision Table: Key Vector Concepts

Additional Information on Vector Operations

Vector operations like the dot product are fundamental in physics and engineering. The dot product of two vectors results in a scalar value, which provides information about the angle between the vectors and their magnitudes. A zero dot product means the vectors are orthogonal (perpendicular). The magnitude of a vector represents its length.

In this problem, the condition that \(\vec{a}+\vec{b}\) is perpendicular to \(\vec{a}\) tells us something specific about the angle between \(\vec{a}\) and \(\vec{b}\). Since \(|\vec{a}|^2 + \vec{a} \cdot \vec{b} = 0\), we have \(\vec{a} \cdot \vec{b} = -|\vec{a}|^2\). Using the definition of the dot product, \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

Substituting the magnitude condition \(|\vec{b}| = 2|\vec{a}|\) and the dot product result \(\vec{a} \cdot \vec{b} = -|\vec{a}|^2\):

\(-|\vec{a}|^2 = |\vec{a}| (2|\vec{a}|) \cos\theta\)

\(-|\vec{a}|^2 = 2|\vec{a}|^2 \cos\theta\)

Assuming \(\vec{a}\) is not the zero vector (which would make the problem trivial, \(|\vec{a}|=0 \implies |\vec{b}|=0\), and \((4\vec{0}+\vec{0})\cdot\vec{0} = 0\)), we can divide by \(|\vec{a}|^2\) (which is non-zero):

\(-1 = 2 \cos\theta\)

\(\cos\theta = -\frac{1}{2}\)

This means the angle \(\theta\) between vectors \(\vec{a}\) and \(\vec{b}\) is \(120^\circ\) or \(2\pi/3\) radians. This additional insight confirms that the conditions given are consistent with a specific geometric arrangement of the vectors.