ข้อสอบ PAT 1 - ตุลาคม 2559

ข้อ 24

กำหนดให้ $\overset{⇀}{a}$ และ $\overset{⇀}{b}$ เป็นเวกเตอร์โดยที่ $\overset{⇀}{a} \cdot \overset{⇀}{b} = 15, |\overset{⇀}{a}| = 6$ และ $(2 \overset{⇀}{a} + \overset{⇀}{b}) \cdot (\overset{⇀}{a} - \overset{⇀}{b}) = 32$

ค่าของ $|\overset{⇀}{a} - 2 \overset{⇀}{b}|$ เท่ากับข้อใดต่อไปนี้

1
$4$
2
$\sqrt{76}$
3
$9$
4
$\sqrt{106}$
5
$\sqrt{136}$

เฉลยข้อสอบ

$จากโจทย์ (2 \overset{⇀}{a} + \overset{⇀}{b}) \cdot (\overset{⇀}{a} - \overset{⇀}{b}) = 32 2 \overset{⇀}{a} \cdot \overset{⇀}{a} - 2 \overset{⇀}{a} \cdot \overset{⇀}{b} + \overset{⇀}{b} \cdot \overset{⇀}{a} - \overset{⇀}{b} \cdot \overset{⇀}{b} = 32$

$โดย \overset{⇀}{a} \cdot \overset{⇀}{a} = {|a|}^{2}, \overset{⇀}{b} \cdot \overset{⇀}{b} = {|b|}^{2}, \overset{⇀}{b} \cdot \overset{⇀}{a} = \overset{⇀}{a} \cdot \overset{⇀}{b} จะได้ 2 {|\overset{⇀}{a}|}^{2} - \overset{⇀}{a} \cdot \overset{⇀}{b} - {|\overset{⇀}{b}|}^{2} = 32; จากโจทย์ \overset{⇀}{a} \cdot \overset{⇀}{b} = 15, |\overset{⇀}{a}| = 6 2 {(6)}^{2} - 15 - {|\overset{⇀}{b}|}^{2} = 32 {|\overset{⇀}{b}|}^{2} = 25; || > 0 เสมอ ดังนั้น |\overset{⇀}{b}| = 5$

$จากสูตร {|\overset{⇀}{u} - \overset{⇀}{v}|}^{2} = {|\overset{⇀}{u}|}^{2} - 2 \overset{⇀}{u} \cdot \overset{⇀}{v} + {|\overset{⇀}{v}|}^{2} จะได้ {|\overset{⇀}{a} - 2 \overset{⇀}{b}|}^{2} = {|\overset{⇀}{a}|}^{2} - 2 \overset{⇀}{a} \cdot (2 \overset{⇀}{b}) + {|2 \overset{⇀}{b}|}^{2}; \overset{⇀}{a} \cdot \overset{⇀}{b} = 15 = {|\overset{⇀}{a}|}^{2} - 4 \overset{⇀}{a} \cdot \overset{⇀}{b} + 4 {|\overset{⇀}{b}|}^{2} = 6^{2} - 4 (15) + 4 (5^{2}) = 76 |\overset{⇀}{a} - 2 \overset{⇀}{b}| = \pm \sqrt{76}; || > 0 เสมอ ดังนั้น |\overset{⇀}{a} - 2 \overset{⇀}{b}| = \sqrt{76}$