Vector Modulus and Dot Product Trick

 

❓ Question

If unit vectors:

$$\vec a,\ \vec b,\ \vec c$$

satisfy:

$$|\vec a-\vec b|^2+|\vec b-\vec c|^2+|\vec c-\vec a|^2=9$$

and

$$|2\vec a+k\vec b+k\vec c|=3$$

then find the positive value of $k$.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Unit vectors
👉 Dot product
👉 Vector identities.

Main idea:

Convert modulus squares into dot products and simplify.

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🔷 Step 1 — Use Unit Vector Property 💯

Since:

$$\vec a,\vec b,\vec c$$

are unit vectors:

$$|\vec a|=|\vec b|=|\vec c|=1$$

Now:

$$|\vec a-\vec b|^2 = |\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b$$

Similarly for all terms:

$$9 = (2-2\vec a\cdot\vec b) + (2-2\vec b\cdot\vec c) + (2-2\vec c\cdot\vec a)$$
$$9 = 6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$$

Thus:

$$\boxed{ \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a = -\frac32 }$$

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🔷 Step 2 — Use Identity

We know:

$$|\vec a+\vec b+\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2 + 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$$

Substitute values:

$$= 3+2\left(-\frac32\right)$$
$$=0$$

Hence:

$$\boxed{ \vec a+\vec b+\vec c=0 }$$

So:

$$\boxed{ \vec b+\vec c=-\vec a }$$

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🔷 Step 3 — Simplify Given Expression

Given:

$$|2\vec a+k\vec b+k\vec c|=3$$

Replace:

$$\vec b+\vec c=-\vec a$$

Then:

$$|2\vec a-k\vec a|=3$$
$$|(2-k)\vec a|=3$$

Since $|\vec a|=1$,

$$|2-k|=3$$

Thus:

$$2-k=\pm3$$

Cases:

Case 1

$$2-k=3$$
$$k=-1$$

Case 2

$$2-k=-3$$
$$k=5$$

Positive value:

$$\boxed{ 5 }$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Unit vector condition use na karna

❌ Identity:

$$|\vec a+\vec b+\vec c|^2$$

miss kar dena

Remember:

Whenever symmetric dot products appear:

$$\boxed{ (\vec a+\vec b+\vec c)^2 }$$

try karna.

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✅ Final Answer

$$\boxed{ 5 }$$

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📚 Related Topics

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• Shortest Distance Between Lines with Parameter α