Question and Answers Forum

All Questions      Topic List

Vector Calculus Questions

Previous in All Question      Next in All Question      

Previous in Vector Calculus      Next in Vector Calculus      

Question Number 56460 by cesar.marval.larez@gmail.com last updated on 16/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19

$$\overrightarrow{w}=3i+4j+5k$$$$\overrightarrow{u}=ai+2j-3k$$$$\overrightarrow{v}=i+4j+ck$$$$\overrightarrow{w}.\overrightarrow{u}=0$$$$(3i+4j+5k).(ai+2j-3k)=0$$$$3a+8-15=0$$$$3a=7\to a=\frac{7}{3}$$$$\overrightarrow{w}.\overrightarrow{v}=0$$$$(3i+4j+5k).(i+4j+ck)=0$$$$3+16+5c=0\to 5c=-19$$$$c=\frac{-19}{5}$$$$b)\stackrel{-}{u}.\overrightarrow{v}=\mid \overrightarrow{u}\mid \mid \overrightarrow{v}\mid cos\theta$$$$cos\theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{\mid \overrightarrow{u}\mid \mid \overrightarrow{v}\mid }=\frac{(ai+2j-3k).(i+4j+ck)}{\sqrt{a^2+4^{}+{(-3)}^2\times \sqrt{1^2+16+c^2}}}$$$$cos\theta =\frac{3a+8-3c}{\sqrt{25+a^2}\}}\times \frac{1}{\sqrt{17+c^2}}=$$$$nowputvalueofandcalculate$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com