If (a + 3b) is perpendicular to (7a - 5b) and (a - 4b) is perpendicular to (7a - 2b), then the angle

Q: Wait, why can't we just assume the vectors are unit vectors? It would make the math so much easier!

That is a tempting shortcut, but it is a dangerous trap! While the final angle is independent of the magnitudes, assuming they are unit vectors is a specific case. By keeping $|\vec{a}|$ and $|\vec{b}|$ as variables, you prove the result holds for *any* non-zero vectors, which is the rigorous way to solve JEE problems.

Q: I got confused with the expansion. Is $\vec{a} \cdot \vec{b}$ the same as $\vec{b} \cdot \vec{a}$?

Absolutely! The dot product is commutative. Just like $3 \times 5 = 5 \times 3$, the scalar product of two vectors is symmetric. This property is exactly what allows us to combine the terms and simplify the equations into a solvable system.

Q: Why did we subtract the two equations? Could we have used substitution instead?

You absolutely could! Substitution is perfectly valid. However, in this specific problem, the coefficients of $|\vec{a}|^2$ are identical in both equations. Subtracting them is a 'surgical strike' that eliminates the $|\vec{a}|^2$ term instantly, saving you time and reducing the chance of arithmetic errors.

Q: How do I know if I should use the dot product or the cross product here?

Great question! The cross product is used when we talk about perpendicularity in terms of planes or areas (since the cross product produces a vector perpendicular to the plane of the two vectors). But when we are given conditions like 'vector A is perpendicular to vector B', we are dealing with the angle between them, which is the domain of the dot product.

Q: Is it possible for the angle to be something other than 60 degrees?

Mathematically, the cosine function is periodic, but the angle between two vectors is defined in the range $[0, 180^\circ]$. Since $\cos \theta = 1/2$ has only one solution in this interval, $60^\circ$ is the unique and definitive answer.

Question

If $(\vec{a} + 3\vec{b})$ is perpendicular to $(7\vec{a} - 5\vec{b})$ and $(\vec{a} - 4\vec{b})$ is perpendicular to $(7\vec{a} - 2\vec{b})$, then the angle between $\vec{a}$ and $\vec{b}$ (in degrees) is ___

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are uncovering the hidden symmetry between two vectors, $\vec{a}$ and $\vec{b}$.

When we look at the condition that $(\vec{a} + 3\vec{b})$ is perpendicular to $(7\vec{a} - 5\vec{b})$, we are observing a geometric constraint that forces these vectors into a specific relationship.

### The Language of Dot Products

In the realm of vectors, 'perpendicular' is the key that unlocks the dot product. Whenever two vectors are perpendicular, their dot product must be zero. Mathematically, if $\vec{u} \perp \vec{v}$, then $\vec{u} \cdot \vec{v} = 0$.

Applying this to our first condition:

$$(\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0$$

By expanding this using the distributive property, we get:

$$7(\vec{a} \cdot \vec{a}) - 5(\vec{a} \cdot \vec{b}) + 21(\vec{b} \cdot \vec{a}) - 15(\vec{b} \cdot \vec{b}) = 0$$

Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, this simplifies to:

$$7|\vec{a}|^2 + 16(\vec{a} \cdot \vec{b}) - 15|\vec{b}|^2 = 0 \quad \dots (1)$$

### The Second Constraint

Now, we repeat this process for the second condition: $(\vec{a} - 4\vec{b}) \perp (7\vec{a} - 2\vec{b})$. Setting the dot product to zero:

$$(\vec{a} - 4\vec{b}) \cdot (7\vec{a} - 2\vec{b}) = 0$$

Expanding this, we obtain:

$$7|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) - 28(\vec{b} \cdot \vec{a}) + 8|\vec{b}|^2 = 0$$

Which simplifies to:

$$7|\vec{a}|^2 - 30(\vec{a} \cdot \vec{b}) + 8|\vec{b}|^2 = 0 \quad \dots (2)$$

### The Elegant Cancellation

Both equations contain the term $7|\vec{a}|^2$. By subtracting equation (2) from equation (1), we eliminate the $|\vec{a}|^2$ term entirely:

$$(7|\vec{a}|^2 + 16\vec{a} \cdot \vec{b} - 15|\vec{b}|^2) - (7|\vec{a}|^2 - 30\vec{a} \cdot \vec{b} + 8|\vec{b}|^2) = 0$$

This yields the relationship:

$$46(\vec{a} \cdot \vec{b}) - 23|\vec{b}|^2 = 0 \Rightarrow \vec{a} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2 \quad \dots (3)$$

### Final Calculation

Now that we know $\vec{a} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2$, we substitute this back into equation (1) to find the relationship between the magnitudes:

$$7|\vec{a}|^2 + 16\left(\frac{1}{2}|\vec{b}|^2ight) - 15|\vec{b}|^2 = 0$$

$$7|\vec{a}|^2 + 8|\vec{b}|^2 - 15|\vec{b}|^2 = 0 \Rightarrow 7|\vec{a}|^2 = 7|\vec{b}|^2 \Rightarrow |\vec{a}| = |\vec{b}|$$

Finally, we use the definition of the dot product, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$:

$$\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\frac{1}{2}|\vec{b}|^2}{|\vec{b}||\vec{b}|} = \frac{1}{2}$$

Since $\cos 	heta = \frac{1}{2}$, we conclude that the angle between the vectors is **$	heta = 60^\circ$**.

Animated Solution for Mathematics - Vector Algebra: If $(a + 3 b)$ is perpendicular to $(7 a - 5 b)$ and $(a - 4 b)$ is perpendicular to $(7 a - 2 b)$ , then the angle between $a$ and $b$ (in degrees) is ___

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

Applying the First Condition

Expanding the First Equation

Applying the Second Condition

Expanding the Second Equation

Eliminating $∣ a ∣^{2}$

Finding $a \cdot b$ in terms of $∣ b ∣^{2}$

Relating $∣ a ∣$ and $∣ b ∣$

Calculating $cos θ$

Final Answer

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Language of Dot Products

The Second Constraint

The Elegant Cancellation

Final Calculation

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