Q.If $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$, then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

Concept: Perpendicular Vectors Condition — two vectors are perpendicular iff their dot product is zero.

Step 1: Compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$.

Step 2: Take the dot product $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$.

Step 3: Since the dot product is zero, the vectors are perpendicular.

Why This Works

The condition for perpendicular vectors is one of the cleanest in vector algebra: if two vectors are at right angles, their dot product equals zero. This is because the dot product measures how much one vector "projects" onto the other — when the projection is zero, the vectors are orthogonal.

Here, we're not given the vectors directly; we're forming them from $\vec{a}$ and $\vec{b}$. The beauty is that $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ have a special relationship — they are like the diagonals of a parallelogram formed by $\vec{a}$ and $\vec{b}$. When $\vec{a}$ and $\vec{b}$ have equal magnitudes, these diagonals are perpendicular. Let's check if that's the case.

Step-by-Step Solution

1. Write down the given vectors clearly.

$\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$

$\vec{b} = \hat{i} + 3\hat{j} - 5\hat{k}$

2. Compute $\vec{a} + \vec{b}$.

Add corresponding components:

• $\hat{i}$: $5 + 1 = 6$
• $\hat{j}$: $-1 + 3 = 2$
• $\hat{k}$: $-3 + (-5) = -8$

So $\vec{a} + \vec{b} = 6\hat{i} + 2\hat{j} - 8\hat{k}$

3. Compute $\vec{a} - \vec{b}$.

Subtract corresponding components:

• $\hat{i}$: $5 - 1 = 4$
• $\hat{j}$: $-1 - 3 = -4$
• $\hat{k}$: $-3 - (-5) = -3 + 5 = 2$

So $\vec{a} - \vec{b} = 4\hat{i} - 4\hat{j} + 2\hat{k}$

4. Take the dot product of these two vectors.

$(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = (6)(4) + (2)(-4) + (-8)(2)$

$= 24 - 8 - 16$

$= 24 - 24 = 0$

5. Interpret the result.

Since the dot product is zero, the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

Method: Proving two constructed vectors are perpendicular

Use this to show combinations such as $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

Steps

Step 1: Recall the orthogonality test.

Two vectors are perpendicular iff their dot product is zero:

Step 2: Form the two vectors, then dot them.

Compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ component-wise (mind the signs when subtracting negative components), then evaluate $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$. A zero result proves perpendicularity.

Step 3: (Shortcut) use the difference-of-squares identity.

so these two are perpendicular exactly when $|\vec{a}|=|\vec{b}|$. Checking the two magnitudes is often faster than forming the sum and difference.

Common Mistakes

Mistake 1: Sign error when subtracting the negative $\hat{k}$-component.

Why it's wrong: $\vec{a}-\vec{b}$ has $\hat{k}$-component $-3-(-5)=+2$, not $-8$; a wrong sign breaks the dot product. Correct approach: subtract carefully, giving $\vec{a}-\vec{b}=4\hat{i}-4\hat{j}+2\hat{k}$, so the dot product is $24-8-16=0$.

Mistake 2: Assuming $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are always perpendicular.

Why it's wrong: $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2$, which is zero only when $|\vec{a}|=|\vec{b}|$. Correct approach: verify equal magnitudes (here both $\sqrt{35}$) — the perpendicularity is a consequence, not automatic.