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ત્રિપરિમાણીય ભૂમિતિ — ગણિત ધોરણ 12 સાયન્સ — Question

Gujarat Boardગુજરાતી માધ્યમધોરણ 12 સાયન્સગણિતત્રિપરિમાણીય ભૂમિતિ1 Mark
MCQ
એક સદિશો $\overrightarrow{a}$ અને $\overrightarrow{b}$ અસમરેખ છે. $\overrightarrow{u}=\overrightarrow{a}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{b}$ અને $\overrightarrow{v}=\overrightarrow{a} \times \overrightarrow{b} $ હોય તો $|\overrightarrow{v}|=\ ........$
  • A
    $|\overrightarrow{u}|+|a|$
  • B
    $|\overrightarrow{u}|+|\overrightarrow{u}.\overrightarrow{a}|$
  • ✓
    $|\overrightarrow{v}|+|\overrightarrow{u}.\overrightarrow{b}|$
  • D
    $\overline{|\overrightarrow{u}|+\overrightarrow{u}.(\overrightarrow{a} + \overrightarrow{b})}$
✓

Answer

Correct option: C.
$|\overrightarrow{v}|+|\overrightarrow{u}.\overrightarrow{b}|$
$|\overrightarrow{a}|=|\overrightarrow{b}|=1$
$\overrightarrow{a}$ અને $\overrightarrow{b}$ અસમરેખ હોવાથી $(\overrightarrow{a},\overrightarrow{b}) 0 , \pi $
ધારોકે $(\overrightarrow{a},^\wedge \overrightarrow{b})=\theta$
હવે ,$\overrightarrow{a}. \overrightarrow{b}=|\overrightarrow{a}\|\overrightarrow{b}|\cos \theta$
$\therefore \overrightarrow{a}. \overrightarrow{b}=\cos \theta$
$\overrightarrow{u}=\overrightarrow{a}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{b}$
$|\overrightarrow{u}|^2=|\overrightarrow{a}|^2-2(\overrightarrow{a}.\overrightarrow{b})(\overrightarrow{a}.\overrightarrow{b})+(\overrightarrow{a}.\overrightarrow{b})^2 \ \ |\overrightarrow{b}|^2$
$=1-2(\overrightarrow{a}.\overrightarrow{b})^2+(\overrightarrow{a}.\overrightarrow{b})^2$
$=1-(\overrightarrow{a}.\overrightarrow{b})^2=1-\cos^2\theta =\sin^2 \theta=|\overrightarrow{a}\times \overrightarrow{b}|^2 $
$(|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}\|\overrightarrow{b}|\sin\theta=\sin\theta)$
$|\overrightarrow{v}|=|\overrightarrow{u}|\ ((\overrightarrow{v}=\overrightarrow{a}\times \overrightarrow{b})$
$\therefore (A)$ સત્ય છે .
હવે,$\overrightarrow{u}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{a}-(\overrightarrow{a}.\overrightarrow{b})(\overrightarrow{a}.\overrightarrow{b})$
$=|\overrightarrow{a}|^2-(\overrightarrow{a}.\overrightarrow{b})^2=1-\cos^2\theta=\sin^2\theta$
$\therefore |\overrightarrow{u}|+|\overrightarrow{u}.\overrightarrow{a}|=|\overrightarrow{v}|+|\overrightarrow{v}^2|$
$|\overrightarrow{v}|$
$\therefore (B)$ સત્ય છે .
$\overrightarrow{u}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})(\overrightarrow{b}.\overrightarrow{b})=\overrightarrow{a}.\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})|\overrightarrow{b}|^2=0$
$|\overrightarrow{v}|+|\overrightarrow{u}.\overrightarrow{b}|=|\overrightarrow{u}|=|\overrightarrow{v}|$
$\therefore (C)$ સત્ય છે .
$\overrightarrow{u}.(\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{u}.\overrightarrow{a}+\overrightarrow{u}.\overrightarrow{b}=|\overrightarrow{v}|^2+\overrightarrow{0}$
$\therefore |\overrightarrow{u}|+\overrightarrow{u}.(\overrightarrow{a}+\overrightarrow{b})=|\overrightarrow{v}|+|\overrightarrow{v}|^2$
$\therefore (D)$ સત્ય નથી .

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