A tricky vector sum

Algebra Level 3

If a ⃗ . b ⃗ + b ⃗ . c ⃗ + c ⃗ . a ⃗ = 0 \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=0 and ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 = k , |\vec{a}+\vec{b}+\vec{c}|^2=k, find the value of l l such that

∣ a ⃗ + b ⃗ − c ⃗ ∣ 2 + ∣ a ⃗ − b ⃗ + c ⃗ ∣ 2 + ∣ − a ⃗ + b ⃗ + c ⃗ ∣ 2 = l ⋅ k . |\vec{a}+\vec{b}-\vec{c}|^2 + |\vec{a}-\vec{b}+\vec{c}|^2+|-\vec{a}+\vec{b}+\vec{c}|^2=l \cdot k.


The answer is 3.

T C Adityaa by T C Adityaa , 18, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

T C Adityaa
Apr 16, 2021

C o n s i d e r [ ( a ⃗ + b ⃗ + c ⃗ ) + ( a ⃗ + b ⃗ − c ⃗ ) ] ⋅ [ ( a ⃗ + b ⃗ + c ⃗ ) − ( a ⃗ + b ⃗ − c ⃗ ) ] . I t r e d u c e s t o 2 ( a ⃗ + b ⃗ ) ⋅ ( 2 c ⃗ ) = 4 ( a ⃗ ⋅ c ⃗ + b ⃗ ⋅ c ) ⟶ e q n 1 H o w e v e r , u s i n g t h e f o r m u l a ( c ⃗ + d ⃗ ) ⋅ ( c ⃗ − d ⃗ ) = ∣ c ⃗ ∣ 2 − ∣ d ⃗ ∣ 2 , w e g e t ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 − ∣ a ⃗ + b ⃗ − c ⃗ ∣ 2 ⟶ e q n 2 E q u a t i n g 1 a n d 2 , w e g e t 4 ( a ⃗ ⋅ c ⃗ + b ⃗ ⋅ c ⃗ ) = ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 − ∣ a ⃗ + b ⃗ − c ⃗ ∣ 2 ⟶ e q n 3 S i m i l a r l y , 4 ( a ⃗ ⋅ b + c ⃗ ⋅ b ⃗ ) = ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 − ∣ a ⃗ − b ⃗ + c ⃗ ∣ 2 ⟶ e q n 4 A n d , 4 ( b ⃗ ⋅ a ⃗ + c ⃗ ⋅ a ⃗ ) = ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 − ∣ − a ⃗ + b ⃗ + c ⃗ ∣ 2 ⟶ e q n 5 A d d i n g e q u a t i o n s 3 , 4 , a n d 5 : 8 ( a ⃗ ⋅ b ⃗ + b ⃗ ⋅ c ⃗ + c ⃗ ⋅ a ⃗ ) = 3 ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 − ( ∣ a ⃗ + b ⃗ − c ⃗ ∣ 2 + ∣ a ⃗ − b ⃗ + c ⃗ ∣ 2 + ∣ − a ⃗ + b ⃗ + c ⃗ ∣ 2 ) A s a ⃗ ⋅ b ⃗ + b ⃗ ⋅ c ⃗ + c ⃗ ⋅ a ⃗ = 0 , w e g e t ∣ a ⃗ + b ⃗ − c ⃗ ∣ 2 + ∣ a ⃗ − b ⃗ + c ⃗ ∣ 2 + ∣ − a ⃗ + b ⃗ + c ⃗ ∣ 2 = 3 ∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 . ∴ l = 3. \begin{array}{l}Consider\ \left[\left(\vec{a}+\vec{b}+\vec{c}\right)+\left(\vec{a}+\vec{b}-\vec{c}\right)\right]\cdot\left[\left(\vec{a}+\vec{b}+\vec{c}\right)-\left(\vec{a}+\vec{b}-\vec{c}\right)\right].\\ It\ reduces\ to\ 2\left(\vec{a}+\vec{b}\right)\cdot\left(2\vec{c}\right)=4\left(\vec{a}\cdot\vec{c}+\vec{b}\cdot c\right)\ \longrightarrow\ eqn\ 1\\ However,\ using\ the\ formula\ \left(\vec{c}+\vec{d}\right)\cdot\left(\vec{c}-\vec{d}\right)=\left|\vec{c}\right|^2-\left|\vec{d}\right|^2,\ we\ get\ \left|\vec{a}+\vec{b}+\vec{c}\right|^2-\left|\vec{a}+\vec{b}-\vec{c}\right|^2\longrightarrow\ eqn\ 2\\ Equating\ 1\ and\ 2,\ we\ get\ 4\left(\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}\right)\ =\ \left|\vec{a}+\vec{b}+\vec{c}\right|^2-\left|\vec{a}+\vec{b}-\vec{c}\right|^2\longrightarrow\ eqn\ 3\\ Similarly,\ 4\left(\vec{a}\cdot b+\vec{c}\cdot\vec{b}\right)\ =\ \left|\vec{a}+\vec{b}+\vec{c}\right|^2-\left|\vec{a}-\vec{b}+\vec{c}\right|^2\longrightarrow\ eqn\ 4\\ And,\ 4\left(\vec{b}\cdot\vec{a}+\vec{c}\cdot\vec{a}\right)\ =\ \left|\vec{a}+\vec{b}+\vec{c}\right|^2-\left|\vec{-a}+\vec{b}+\vec{c}\right|^2\longrightarrow\ eqn\ 5\\ \\ Adding\ equations\ 3,4,\ and\ 5:\\ \\ 8\left(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}\right)=3\left|\vec{a}+\vec{b}+\vec{c}\right|^2-\left(\left|\vec{a}+\vec{b}-\vec{c}\right|^2+\left|\vec{a}-\vec{b}+\vec{c}\right|^2+\left|\vec{-a}+\vec{b}+\vec{c}\right|^2\right)\\ As\ \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=0,\ we\ get\ \left|\vec{a}+\vec{b}-\vec{c}\right|^2+\left|\vec{a}-\vec{b}+\vec{c}\right|^2+\left|\vec{-a}+\vec{b}+\vec{c}\right|^2\ =\ 3\left|\vec{a}+\vec{b}+\vec{c}\right|^2.\\ \therefore l=3.\\ \end{array}

1 Helpful 1 Interesting 2 Brilliant 0 Confused

Just expanding the terms in the L.H.S of the question results in 27 terms. Not quite an efficient method. Also, just substituting suitable values like i, j and k is not that satisfying of a solution

T C Adityaa - 1 month, 3 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...