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Vector Equation for Straight Line

Given, two ve...

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Given, two vectors A=-4i+4j+2k and B=2i-j-k. The angle made by (A+B) with i+2j-4k is

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Q. Find the angle between the vectors

\vec{a} and \vec{b}

\vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{j} + \hat{k}

\vec{a} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k} and \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}

\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = 4 \hat{i} + 4 \hat{j} - 2 \hat{k}

\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}

\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}

\to A = 2 ˆ i + 4 ˆ j + 4 ˆ k

\to B = 4 ˆ i + 2 ˆ j - 4 ˆ k

are two vectors. The angle between them will be

\vec{a} \cdot \vec{b}

\vec{a} = \hat{i} - 2 \hat{j} + \hat{k} and \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k}

\vec{a} = \hat{j} + 2 \hat{k} and \vec{b} = 2 \hat{i} + \hat{k}

\vec{a} = \hat{j} - \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k}

Q. Show that the points whose position vectors are as given below are collinear:
(i)

2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k} and \hat{i} + 4 \hat{j} - 3 \hat{k}

3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k} and - \hat{i} + 4 \hat{j} - 2 \hat{k}

Q. Show the each of the following triads of vectors are coplanar:
(i)

\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} + 2 \hat{j} + 7 \hat{k}, \vec{c} = 5 \hat{i} + 6 \hat{j} + 5 \hat{k}

\vec{a} = - 4 \hat{i} - 6 \hat{j} - 2 \hat{k}, \vec{b} = - \hat{i} + 4 \hat{j} + 3 \hat{k}, \vec{c} = - 8 \hat{i} - \hat{j} + 3 \hat{k}

\hat{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \hat{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \hat{c} = \hat{i} - 3 \hat{j} + 5 \hat{k}

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