Vectors

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Draw a horizontal line (x-axis). From a point, draw a vector inclined at 30° above the x-axis. Label its length proportional to 50 N. The vector has magnitude 50 N and direction 30° from horizontal.

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Magnitude \( |\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). So, the length of the vector is 5 units.

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Step 1: Find magnitude \( |\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \). Step 2: Unit vector \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2}{7}\hat{i} - \frac{3}{7}\ha…

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Add component-wise: \( (1+2)\hat{i} + (2-1)\hat{j} + (-1+3)\hat{k} = 3\hat{i} + \hat{j} + 2\hat{k} \).

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Subtract component-wise: \( (4-1)\hat{i} + (-1-3)\hat{j} = 3\hat{i} - 4\hat{j} \).

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Multiply each component: \( 3 \times 2\hat{i} = 6\hat{i} \), \( 3 \times (-5)\hat{j} = -15\hat{j} \). So, \( 3\vec{v} = 6\hat{i} - 15\hat{j} \).

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Yes, they are collinear. Because \( \vec{b} = 2\vec{a} \), so one is a scalar multiple of the other, meaning they lie on the same line.

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Let A, B, C have position vectors. Then \( \overrightarrow{AB} = (3\hat{i}+4\hat{j}) - (2\hat{i}+3\hat{j}) = \hat{i} + \hat{j} \). \( \overrightarrow{BC} = (5\hat{i}+6\hat{j}) - (3\hat{i}+4\hat{j}) = …