Q.Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})$, $\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$, $\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ Also, show that they are mutually perpendicular to each other.

Concept: Unit Vector Verification & Orthogonality

A vector is a unit vector if its magnitude equals 1. Two vectors are perpendicular if their dot product is zero.

Step 1: Magnitude of the first vector

Step 2: Magnitudes of the second and third vectors

By symmetry, the second vector $\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$ has magnitude $\frac{1}{7}\sqrt{9+36+4}=1$, and the third $\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ has magnitude $\frac{1}{7}\sqrt{36+4+9}=1$. All three are unit vectors.

Step 3: Dot products to check perpendicularity

• First · Second: $\frac{1}{49}(2\cdot3 + 3\cdot(-6) + 6\cdot2) = \frac{1}{49}(6-18+12)=0$ …

Why this works

A unit vector is any vector whose length (magnitude) equals exactly 1. To check if a vector is a unit vector, you compute its magnitude — if it’s 1, you’re done. For a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, the magnitude is $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.

Two vectors are perpendicular (orthogonal) if their dot product is zero. The dot product of $\vec{a}$ and $\vec{b}$ is $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$. If this equals 0, the vectors are at right angles.

Here, each vector is given as a scalar multiple $\frac{1}{7}$ times a triplet of integers. That factor $\frac{1}{7}$ is chosen deliberately — the integer triplets are Pythagorean triples in 3D, so their squared sums are perfect squares. Let’s verify.

Step-by-step verification

Let’s name the three vectors:

1. Check that $\vec{u}$ is a unit vector

Magnitude squared:

So $|\vec{u}| = 1$. It’s a unit vector.

2. Check $\vec{v}$

Unit vector.

3. Check $\vec{w}$

Unit vector.

4. Show $\vec{u}$ and $\vec{v}$ are perpendicular

Dot product:

Perpendicular.

5. Show $\vec{u}$ and $\vec{w}$ are perpendicular …

Method: Verifying Unit Vectors and Mutual Perpendicularity

Use this when asked to show a set of vectors are each unit vectors and are mutually perpendicular (an orthonormal set).

Steps

Step 1: Test each for unit length.

A vector is a unit vector iff its magnitude is $1$. For a scaled vector $\tfrac1k(p\hat i+q\hat j+r\hat k)$:

Show this equals $1$ (i.e. $\sqrt{p^2+q^2+r^2}=k$).

Step 2: Test each pair for perpendicularity.

Two vectors are perpendicular iff their dot product is $0$:

$\vec{u}\cdot\vec{v}=0.$ …

Common Mistakes

Mistake 1: Ignoring the $\tfrac1k$ factor when finding magnitude.

Why it's wrong: the length is $\tfrac{1}{7}\sqrt{p^2+q^2+r^2}$; skipping the $\tfrac17$ makes it look like $\sqrt{49}=7$ instead of $1$. Correct approach: divide the root by $k=7$ (or check $\sqrt{p^2+q^2+r^2}=7$).

Mistake 2: Checking only one or two of the three pairs.

Why it's wrong: "mutually perpendicular" requires every pair orthogonal, so all three dot products must vanish. Correct approach: test $\vec u\!\cdot\!\vec v$, $\vec u\!\cdot\!\vec w$, and $\vec v\!\cdot\!\vec w$. …