A geometry problem by Steven Chase

Geometry Level 3

v 1 ⃗ = 1 ı ^ + 3 ȷ ^ v 2 ⃗ = 2 ı ^ + 1 ȷ ^ v 3 ⃗ = ? \vec{v_1} = 1 \hat{\imath} + 3 \hat{\jmath} \\ \vec{v_2} = 2 \hat{\imath} + 1 \hat{\jmath} \\ \vec{v_3} = ?

Define a resultant vector as follows:

v r e s ⃗ = v 1 ⃗ + v 2 ⃗ + v 3 ⃗ \vec{v_{res}} = \vec{v_1} + \vec{v_2} + \vec{v_3}

What is the shortest possible length of v 3 ⃗ \vec{v_3} , such that the length of v r e s ⃗ \vec{v_{res}} is equal to 3?


The answer is 2.

Steven Chase by Steven Chase , 37, USA

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.

2 solutions

Saya Suka
Dec 21, 2016

Shortest length.
= √[(1+2)^2+(3+1)^2] - 3
= 5 - 3.
= 2


3 Helpful 0 Interesting 0 Brilliant 0 Confused

this problem is very cool! :)

Charlotte Milanese - 4 years, 5 months ago
Kishore S. Shenoy
Dec 25, 2016

First of all, the triangle inequality, ∣ ∣ z 1 ⃗ ∣ − ∣ z 2 ⃗ ∣ ∣ ≤ ∣ z 1 ⃗ ± z 2 ⃗ ∣ ≤ ∣ z 1 ⃗ ∣ + ∣ z 2 ⃗ ∣ \left||\vec {z_1}|-|\vec {z_2}|\right|\le\left|\vec{ z_1}\pm\vec{ z_2}\right|\le|\vec {z_1}|+|\vec {z_2}|

Now, I'm writing v 3 ⃗ = v res ⃗ − ( v 1 ⃗ + v 2 ⃗ ) = v res ⃗ − ( 3 ı ^ + 4 ȷ ^ ) \vec{v_3} = \vec{v_{\text{res}}} - \left(\vec{v_1}+\vec{v_2}\right) =\vec{v_{\text{res}}} - \left(3\hat {\imath} + 4\hat{\jmath}\right)

Using the first part of triangle inequality, v 3 ⃗ ≥ ∣ ∣ v res ⃗ ∣ − ∣ 3 ı ^ + 4 ȷ ^ ∣ ∣ = 5 − 3 = 2 \vec{v_3}\ge \left||\vec {v_{\text{res}}}|-|3\hat {\imath} + 4\hat{\jmath}|\right| = 5-3 = \color{#3D99F6}{\boxed{2}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...