मान लीजिए vec alpha = 3hat i + hat j और vec b | vedclass

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(A) दिया गया है $\vec{\alpha} = 3\hat{i} + \hat{j}$ और $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}.$चूंकि $\vec{\beta}_1$ सदिश $\vec{\alpha}$ के समां

Vedclass · Accepted Answer

$\frac{1}{2}(-3\hat i + 9\hat j + 5\hat k)$

Vedclass · Answer

(A) दिया गया है $\vec{\alpha} = 3\hat{i} + \hat{j}$ और $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}.$चूंकि $\vec{\beta}_1$ सदिश $\vec{\alpha}$ के समांतर है,इसलिए $\vec{\beta}_1 = \lambda \vec{\alpha} = \lambda(3\hat{i} + \hat{j}).$दिया गया है $\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2,$ इसलिए $\vec{\beta}_2 = \vec{\beta}_1 - \vec{\beta} = \lambda(3\hat{i} + \hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k}) = (3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}.$चूंकि $\vec{\beta}_2$ सदिश $\vec{\alpha}$ के लंबवत है,इसलिए $\vec{\beta}_2 \cdot \vec{\alpha} = 0.$$((3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k}) \cdot (3\hat{i} + \hat{j}) = 0.$$3(3\lambda - 2) + 1(\lambda + 1) = 0 \implies 9\lambda - 6 + \lambda + 1 = 0 \implies 10\lambda = 5 \implies \lambda = \frac{1}{2}.$अतः,$\vec{\beta}_1 = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}$ और $\vec{\beta}_2 = (\frac{3}{2} - 2)\hat{i} + (\frac{1}{2} + 1)\hat{j} - 3\hat{k} = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}.$अब,$\vec{\beta}_1 	imes \vec{\beta}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \frac{3}{2} & \frac{1}{2} & 0 \ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix}.$$= \hat{i}(-\frac{3}{2} - 0) - \hat{j}(-\frac{9}{2} - 0) + \hat{k}(\frac{9}{4} - (-\frac{1}{4})) = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{10}{4}\hat{k} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k}.$$= \frac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k}).$

Similar Questions

यदि $\overrightarrow{a}+2 \overrightarrow{b}+3 \overrightarrow{c}=\overrightarrow{0}$ है,तो $\overrightarrow{a} \times \overrightarrow{b}+\overrightarrow{b} \times \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}$ का मान ज्ञात कीजिए।

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