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Miscellaneous Exercise · Q5

Q.Find the value of xx for which x(i^+j^+k^)x(\hat{i} + \hat{j} + \hat{k}) is a unit vector.

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A unit vector has magnitude 1. For x(i^+j^+k^)x(\hat{i} + \hat{j} + \hat{k}) to be a unit vector, we set its magnitude equal to 1 and solve for xx, giving x=±13x = \pm \frac{1}{\sqrt{3}}.

Why This Works: The Unit Vector Idea

A unit vector is any vector whose length (magnitude) is exactly 1. The vector given here is xx multiplied by (i^+j^+k^)(\hat{i} + \hat{j} + \hat{k}). So the question is: what value of xx scales this particular direction vector so that its final length becomes 1?

The key insight: multiplying a vector by a scalar xx scales its magnitude by ∣x∣|x|. If we know the magnitude of (i^+j^+k^)(\hat{i} + \hat{j} + \hat{k}), we can find the xx that makes the product's magnitude equal to 1.

For any vector v⃗=ai^+bj^+ck^\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}, its magnitude is ∣v⃗∣=a2+b2+c2|\vec{v}| = \sqrt{a^2 + b^2 + c^2}.

Step-by-Step Solution

1. Find the magnitude of the base vector i^+j^+k^\hat{i} + \hat{j} + \hat{k}.

Each component is 1. So:

∣i^+j^+k^∣=12+12+12=3|\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}

2. Understand how scalar multiplication affects magnitude.

If v⃗=i^+j^+k^\vec{v} = \hat{i} + \hat{j} + \hat{k}, then xv⃗x\vec{v} has magnitude:

∣xv⃗∣=∣x∣⋅∣v⃗∣=∣x∣⋅3|x\vec{v}| = |x| \cdot |\vec{v}| = |x| \cdot \sqrt{3}

Watch out

A common mistake is to forget the absolute value. The magnitude of xv⃗x\vec{v} is ∣x∣⋅∣v⃗∣|x| \cdot |\vec{v}|, not x⋅∣v⃗∣x \cdot |\vec{v}|. This is why xx can be negative — the magnitude is always non-negative.

3. Set the magnitude equal to 1 and solve.

We want ∣xv⃗∣=1|x\vec{v}| = 1, so: …

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