Question #b00d1

Question

Question #b00d1

1 Answer

Impact of this question

2861 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-03T17:50:25

$| b + a | = \sqrt{3}$ $abs(b+a) = sqrt(3)$

Explanation:

Consider the diagram below.
$\to b = - - \to P Q$ $vec(b)=vec(PQ)$ and $∣ ∣ ∣ - - \to P Q ∣ ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = 1$ $abs(vec(PQ))=abs(vecb)=1$

$\to a = - - \to P Q = - \to T S$ $veca = vec(PQ)=vec(TS)$ and $∣ ∣ ∣ - - \to P Q ∣ ∣ ∣ = ∣ ∣ ∣ - \to T S ∣ ∣ ∣ = ∣ ∣ \to a ∣ ∣ = 1$ $abs(vec(PQ)) = abs(vec(TS)) = abs(veca)=1$

$- -- \to a + b = - \to P S$ $vec(a+b) = vec(PS)$

$∠ S T U = ∠ Q P T = 60^{\circ}$ $/_STU = /_QPT = 60^@$ (the angle between $\to a$ $vec(a)$ and $\to b$ $vec(b)$
$\Rightarrow ∠ P T S = 120^{\circ}$ $rArr /_PTS = 120^@$
$\Rightarrow ∠ R P T = ∠ R S T = 30^{\circ}$ $rArr /_RPT =/_RST=30^@$

The bisector of $∠ P T S$ $/_PTS$
divides $△ P T S$ $triangle PTS$ into 2 congruent triangles
$| P R | = | R S |$ $abs(PR) = abs(RS)$

Since $∠ P T R = 60^{\circ}$ $/_PTR=60^@$ and $| P T | = 1$ $abs(PT)=1$
$\Rightarrow | P R | = \frac{\sqrt{3}}{2}$ $rArr abs(PR) = sqrt(3)/2$

and
$| P S | = \sqrt{3}$ $abs(PS) = sqrt(3)$

Therefore $∣ ∣ ∣ - -- \to a + b ∣ ∣ ∣ = \sqrt{3}$ $abs(vec(a+b)) = sqrt(3)$

enter image source here

Science

Math

Humanities

... and beyond

Question #b00d1

1 Answer

Explanation:

Related questions

Impact of this question