Question:
The angle between the lines $\vec{r}=(3+\alpha)\hat{i}+2(1+\alpha)\hat{j}+2(-2+\alpha)\hat{k}$ and $\vec{r}=(5+3\beta)\hat{i}+2(1+\beta)\hat{j}+6\beta\hat{k}$ , where $\alpha$ and $\beta$ are parameters, is
The angle between the lines $\vec{r}=(3+\alpha)\hat{i}+2(1+\alpha)\hat{j}+2(-2+\alpha)\hat{k}$ and $\vec{r}=(5+3\beta)\hat{i}+2(1+\beta)\hat{j}+6\beta\hat{k}$ , where $\alpha$ and $\beta$ are parameters, is
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Logic Tip: The coefficients of the parameters ($\alpha$ and $\beta$) represent the components of the direction vectors directly. Extracting them mentally saves writing out the fully separated vector equation.
Updated On: Apr 27, 2026
- $\cos^{-1}\left(\frac{17}{21}\right)$
- $\cos^{-1}\left(\frac{19}{12}\right)$
- $\cos^{-1}\left(\frac{9}{21}\right)$
- $2\cos^{-1}\left(\frac{19}{21}\right)$
- $\cos^{-1}\left(\frac{19}{21}\right)$
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Solution and Explanation
Concept:
The angle $\theta$ between two lines given in vector form $\vec{r} = \vec{a} + \lambda\vec{m}$ is determined entirely by their direction vectors ($\vec{m}_1$ and $\vec{m}_2$). The formula is: $$\cos \theta = \frac{\vec{m}_1 \cdot \vec{m}_2}{|\vec{m}_1| |\vec{m}_2|}$$
Step 1: Extract the direction vectors from the line equations.
For Line 1: $$\vec{r} = (3 + \alpha)\hat{i} + (2 + 2\alpha)\hat{j} + (-4 + 2\alpha)\hat{k}$$ Separate the position part and the parameter $\alpha$ part: $$\vec{r} = (3\hat{i} + 2\hat{j} - 4\hat{k}) + \alpha(\hat{i} + 2\hat{j} + 2\hat{k})$$ Thus, the direction vector is $\vec{m}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$. For Line 2: $$\vec{r} = (5 + 3\beta)\hat{i} + (2 + 2\beta)\hat{j} + 6\beta\hat{k}$$ $$\vec{r} = (5\hat{i} + 2\hat{j}) + \beta(3\hat{i} + 2\hat{j} + 6\hat{k})$$ Thus, the direction vector is $\vec{m}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
Step 2: Calculate the dot product of the direction vectors.
$$\vec{m}_1 \cdot \vec{m}_2 = (1)(3) + (2)(2) + (2)(6)$$ $$\vec{m}_1 \cdot \vec{m}_2 = 3 + 4 + 12 = 19$$
Step 3: Calculate the magnitudes of the direction vectors.
$$|\vec{m}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$ $$|\vec{m}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$
Step 4: Calculate the angle $\theta$.
$$\cos \theta = \frac{19}{3 \cdot 7} = \frac{19}{21}$$ $$\theta = \cos^{-1}\left(\frac{19}{21}\right)$$
The angle $\theta$ between two lines given in vector form $\vec{r} = \vec{a} + \lambda\vec{m}$ is determined entirely by their direction vectors ($\vec{m}_1$ and $\vec{m}_2$). The formula is: $$\cos \theta = \frac{\vec{m}_1 \cdot \vec{m}_2}{|\vec{m}_1| |\vec{m}_2|}$$
Step 1: Extract the direction vectors from the line equations.
For Line 1: $$\vec{r} = (3 + \alpha)\hat{i} + (2 + 2\alpha)\hat{j} + (-4 + 2\alpha)\hat{k}$$ Separate the position part and the parameter $\alpha$ part: $$\vec{r} = (3\hat{i} + 2\hat{j} - 4\hat{k}) + \alpha(\hat{i} + 2\hat{j} + 2\hat{k})$$ Thus, the direction vector is $\vec{m}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$. For Line 2: $$\vec{r} = (5 + 3\beta)\hat{i} + (2 + 2\beta)\hat{j} + 6\beta\hat{k}$$ $$\vec{r} = (5\hat{i} + 2\hat{j}) + \beta(3\hat{i} + 2\hat{j} + 6\hat{k})$$ Thus, the direction vector is $\vec{m}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
Step 2: Calculate the dot product of the direction vectors.
$$\vec{m}_1 \cdot \vec{m}_2 = (1)(3) + (2)(2) + (2)(6)$$ $$\vec{m}_1 \cdot \vec{m}_2 = 3 + 4 + 12 = 19$$
Step 3: Calculate the magnitudes of the direction vectors.
$$|\vec{m}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$ $$|\vec{m}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$
Step 4: Calculate the angle $\theta$.
$$\cos \theta = \frac{19}{3 \cdot 7} = \frac{19}{21}$$ $$\theta = \cos^{-1}\left(\frac{19}{21}\right)$$
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