જો બિંદુઓ A, B, C, D ના સ્થાન સદિશો અનુક્રમે | vedclass

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(D) આપેલ સ્થાન સદિશો:$A = \hat{i}+2 \hat{j}+3 \hat{k}$$B = 2 \hat{i}-\hat{j}+2 \hat{k}$$C = \frac{7}{4} \hat{i}+\frac{15}{4} \hat{j}+\frac{15}{4} \hat

Vedclass · Accepted Answer

$187$

Vedclass · Answer

(D) આપેલ સ્થાન સદિશો:$A = \hat{i}+2 \hat{j}+3 \hat{k}$$B = 2 \hat{i}-\hat{j}+2 \hat{k}$$C = \frac{7}{4} \hat{i}+\frac{15}{4} \hat{j}+\frac{15}{4} \hat{k}$$D = \frac{7}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{5+3a}{3} \hat{k}$સદિશ $\vec{AC} = C - A = (\frac{7}{4}-1) \hat{i} + (\frac{15}{4}-2) \hat{j} + (\frac{15}{4}-3) \hat{k} = \frac{3}{4} \hat{i} + \frac{7}{4} \hat{j} + \frac{3}{4} \hat{k}$.સદિશ $\vec{BD} = D - B = (\frac{7}{3}-2) \hat{i} + (\frac{2}{3}-(-1)) \hat{j} + (\frac{5+3a}{3}-2) \hat{k} = \frac{1}{3} \hat{i} + \frac{5}{3} \hat{j} + \frac{3a-1}{3} \hat{k}$.આપેલ છે કે $|AC| = |BD|$,તેથી $|AC|^2 = |BD|^2$:$(\frac{3}{4})^2 + (\frac{7}{4})^2 + (\frac{3}{4})^2 = (\frac{1}{3})^2 + (\frac{5}{3})^2 + (\frac{3a-1}{3})^2$$\frac{9+49+9}{16} = \frac{1+25+(3a-1)^2}{9}$$\frac{67}{16} = \frac{26+(3a-1)^2}{9}$$603 = 16(26 + (3a-1)^2)$$603 = 416 + 16(3a-1)^2$$16(3a-1)^2 = 603 - 416 = 187$.

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વાસ્તવિક સંખ્યાઓની ત્રિપુટી $(x, y, z)$ એવી રીતે કે જેથી $(3 \hat{i}-\hat{j}+2 \hat{k})=(2 \hat{i}+3 \hat{j}-\hat{k}) x+(\hat{i}-2 \hat{j}+2 \hat{k}) y+(-2 \hat{i}+\hat{j}-2 \hat{k}) z$ થાય,તે શોધો.

જો $\bar{a}=\bar{i}-2\bar{j}+2\bar{k}$,$\bar{b}=6\bar{i}+3\bar{j}-2\bar{k}$,અને $\bar{c}=-4\bar{i}+3\bar{j}+12\bar{k}$ ત્રણ સદિશો હોય,તો $\sqrt{(|\bar{a}|+|\bar{b}|+|\bar{c}|)+|\bar{a}+\bar{b}+\bar{c}|}=$

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