ધારો કે vec{a}=4 hat{i}-hat{j}+hat{k},vec{b}= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) આપેલ છે કે $(\vec{a}+\vec{b}) 	imes \vec{c} = \vec{c} 	imes (-2 \vec{a}+3 \vec{b})$.આને $(\vec{a}+\vec{b}) 	imes \vec{c} + (-2 \vec{a}+3 \vec{b

Vedclass · Accepted Answer

$1618$

Vedclass · Answer

(B) આપેલ છે કે $(\vec{a}+\vec{b}) 	imes \vec{c} = \vec{c} 	imes (-2 \vec{a}+3 \vec{b})$.આને $(\vec{a}+\vec{b}) 	imes \vec{c} + (-2 \vec{a}+3 \vec{b}) 	imes \vec{c} = \vec{0}$ તરીકે લખી શકાય.$(\vec{a}+\vec{b}-2 \vec{a}+3 \vec{b}) 	imes \vec{c} = \vec{0}$.$(4 \vec{b}-\vec{a}) 	imes \vec{c} = \vec{0}$.આનો અર્થ એ છે કે $\vec{c}$ એ $(4 \vec{b}-\vec{a})$ ને સમાંતર છે.ધારો કે $\vec{c} = \lambda(4 \vec{b}-\vec{a})$.$4 \vec{b}-\vec{a} = 4(11 \hat{i}-\hat{j}+\hat{k}) - (4 \hat{i}-\hat{j}+\hat{k}) = 40 \hat{i}-3 \hat{j}+3 \hat{k}$.તેથી,$\vec{c} = \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})$.આપેલ છે કે $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$.$2 \vec{a}+3 \vec{b} = 2(4 \hat{i}-\hat{j}+\hat{k}) + 3(11 \hat{i}-\hat{j}+\hat{k}) = 41 \hat{i}-5 \hat{j}+5 \hat{k}$.$(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k}) = 1670$.$\lambda(1640 + 15 + 15) = 1670$.$1670 \lambda = 1670 \Rightarrow \lambda = 1$.આમ,$\vec{c} = 40 \hat{i}-3 \hat{j}+3 \hat{k}$.$|\vec{c}|^2 = 40^2 + (-3)^2 + 3^2 = 1600 + 9 + 9 = 1618$.

Similar Questions

$A, B, C, D$ કોઈપણ ચાર બિંદુઓ હોય,તો $\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = $

$\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ નો $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$ પરનો લંબ પ્રક્ષેપ સદિશ શોધો.

$\vec{i} + \vec{j}$ નો $\vec{j} + \vec{k}$ ની દિશામાં ઘટક શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module