If m_1, m_2, m_3 and m_4 are respectively the | vedclass

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(A) Given vectors are $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\h

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$m_3 < m_1 < m_4 < m_2$

Vedclass · Answer

(A) Given vectors are $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$, and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$.Calculating the magnitudes $m_1, m_2, m_3, m_4$:$m_1 = |\overrightarrow{a}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$$m_2 = |\overrightarrow{a}_2| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9+16+16} = \sqrt{41}$$m_3 = |\overrightarrow{a}_3| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1+1+1} = \sqrt{3}$$m_4 = |\overrightarrow{a}_4| = \sqrt{(-1)^2 + 3^2 + 1^2} = \sqrt{1+9+1} = \sqrt{11}$Comparing the values: $\sqrt{3} Therefore, $m_3 < m_1 < m_4 < m_2$.

Similar Questions

Let the vectors $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ be two sides of a triangle $ABC$. If $G$ is the centroid of $\triangle ABC$, then $\frac{27}{7}(\overrightarrow{AG})^2 + 5 =$

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are parallel vectors,then which of the following is true?

If $a = \hat{i} + 4 \hat{j}$, $b = 2 \hat{i} - 2 \hat{j}$, and $c = 5 \hat{i} + 9 \hat{j}$, then $c$ is equal to:

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.

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