Core Logic
Given a, b, c$\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so | a|² = | b|² = | c|² = 1$|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$.
Expand the given equation | a- b|²+| b- c|²+| c- a|²=9$|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$:
2| a|² - 2 a· b + 2| b|² - 2 b· c + 2| c|² - 2 c· a = 9$$ 2|\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + 2|\vec{b}|^2 - 2\vec{b}\cdot\vec{c} + 2|\vec{c}|^2 - 2\vec{c}\cdot\vec{a} = 9 $$
2(1) - 2 a· b + 2(1) - 2 b· c + 2(1) - 2 c· a = 9$$ 2(1) - 2\vec{a}\cdot\vec{b} + 2(1) - 2\vec{b}\cdot\vec{c} + 2(1) - 2\vec{c}\cdot\vec{a} = 9 $$
6 - 2( a· b + b· c + c· a) = 9$$ 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9 $$
a· b + b· c + c· a = -(3)/(2)$$ \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2} $$
Step 1: Determine Vector Sum
We also know the identity:
| a + b + c|² = | a|² + | b|² + | c|² + 2( a· b + b· c + c· a)$$ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) $$
Substitute the known values:
| a + b + c|² = 1 + 1 + 1 + 2(-(3)/(2)) = 3 - 3 = 0$$ |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 2\left(-\frac{3}{2}\right) = 3 - 3 = 0 $$
Therefore, a + b + c = 0$\vec{a} + \vec{b} + \vec{c} = \vec{0}$, which means b + c = - a$\vec{b} + \vec{c} = -\vec{a}$.
Step 2: Find k
We are given |2 a + k b + k c| = 3$|2\vec{a} + k\vec{b} + k\vec{c}| = 3$.
Rewrite it as:
|2 a + k( b + c)| = 3$$ |2\vec{a} + k(\vec{b} + \vec{c})| = 3 $$
Substitute b + c = - a$\vec{b} + \vec{c} = -\vec{a}$:
|2 a + k(- a)| = 3$$ |2\vec{a} + k(-\vec{a})| = 3 $$
|(2 - k) a| = 3$$ |(2 - k)\vec{a}| = 3 $$
|2 - k| | a| = 3$$ |2 - k| |\vec{a}| = 3 $$
Since | a| = 1$|\vec{a}| = 1$:
|2 - k| = 3$|2 - k| = 3$
This gives two equations:
2 - k = 3 k = -1$2 - k = 3 \implies k = -1$
2 - k = -3 k = 5$2 - k = -3 \implies k = 5$
Step 3: Conclusion
The question asks for the positive value of k$k$, so k = 5$k = 5$.
Chapter Mix
Class 12 Mathematics: Vector Algebra
