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(i) The position vectors of two points A...

(i) The position vectors of two points A and B are `3 hati + hatj + 2 hatk and hati - 2 hatj - 4 hatk` respectively. Find the vector equation of the plane passing throug B and perpendicular to `vec(AB)`.
(ii) Find the vector equation of the plane through the point (2,0, -1) and perpendicular to the line joining the two points (1,2,3,) and (3, -1, 6).

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Let's solve the given questions step by step. ### Part (i) **Given:** - Position vector of point A: \(\vec{A} = 3\hat{i} + \hat{j} + 2\hat{k}\) - Position vector of point B: \(\vec{B} = \hat{i} - 2\hat{j} - 4\hat{k}\) **Step 1: Find the vector \(\vec{AB}\)** The vector \(\vec{AB}\) can be calculated as: \[ \vec{AB} = \vec{A} - \vec{B} \] Substituting the values: \[ \vec{AB} = (3\hat{i} + \hat{j} + 2\hat{k}) - (\hat{i} - 2\hat{j} - 4\hat{k}) \] \[ = (3 - 1)\hat{i} + (1 + 2)\hat{j} + (2 + 4)\hat{k} \] \[ = 2\hat{i} + 3\hat{j} + 6\hat{k} \] **Hint:** To find the vector between two points, subtract the position vector of the second point from the first point. --- **Step 2: Find the normal vector of the plane** The normal vector \(\vec{n}\) of the plane is the same as the vector \(\vec{AB}\): \[ \vec{n} = 2\hat{i} + 3\hat{j} + 6\hat{k} \] **Hint:** The direction of the normal vector of the plane is given by the vector that is perpendicular to the plane. --- **Step 3: Write the vector equation of the plane** The vector equation of the plane can be given by: \[ (\vec{r} - \vec{B}) \cdot \vec{n} = 0 \] Where \(\vec{r}\) is the position vector of a general point on the plane. Substituting \(\vec{B}\) and \(\vec{n}\): \[ (\vec{r} - (\hat{i} - 2\hat{j} - 4\hat{k})) \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 0 \] **Hint:** The equation of a plane can be expressed using the dot product of the normal vector and the vector from a point on the plane to a general point. --- **Step 4: Expand the equation** Expanding the equation: \[ (\vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k})) - ((\hat{i} - 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 6\hat{k})) = 0 \] Calculating the dot product: \[ (\hat{i} - 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 2 - 6 - 24 = -28 \] Thus, the equation becomes: \[ \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) + 28 = 0 \] **Final Equation of the Plane:** \[ \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = -28 \] --- ### Part (ii) **Given:** - Point \(P(2, 0, -1)\) - Points \(A(1, 2, 3)\) and \(B(3, -1, 6)\) **Step 1: Find the direction vector of line AB** The direction vector \(\vec{AB}\) is given by: \[ \vec{AB} = \vec{B} - \vec{A} = (3 - 1)\hat{i} + (-1 - 2)\hat{j} + (6 - 3)\hat{k} \] \[ = 2\hat{i} - 3\hat{j} + 3\hat{k} \] **Hint:** The direction vector of a line can be found by subtracting the position vectors of the two points. --- **Step 2: Write the normal vector of the plane** The normal vector \(\vec{n}\) of the plane is the same as the direction vector \(\vec{AB}\): \[ \vec{n} = 2\hat{i} - 3\hat{j} + 3\hat{k} \] **Hint:** The normal vector of the plane is perpendicular to the line joining the two points. --- **Step 3: Write the vector equation of the plane** The vector equation of the plane can be given by: \[ (\vec{r} - \vec{P}) \cdot \vec{n} = 0 \] Where \(\vec{P} = 2\hat{i} + 0\hat{j} - \hat{k}\). Substituting \(\vec{P}\) and \(\vec{n}\): \[ (\vec{r} - (2\hat{i} + 0\hat{j} - \hat{k})) \cdot (2\hat{i} - 3\hat{j} + 3\hat{k}) = 0 \] **Hint:** The equation of a plane can also be expressed using a known point and a normal vector. --- **Step 4: Expand the equation** Expanding the equation: \[ (\vec{r} \cdot (2\hat{i} - 3\hat{j} + 3\hat{k})) - ((2\hat{i} + 0\hat{j} - \hat{k}) \cdot (2\hat{i} - 3\hat{j} + 3\hat{k})) = 0 \] Calculating the dot product: \[ (2\hat{i} + 0\hat{j} - \hat{k}) \cdot (2\hat{i} - 3\hat{j} + 3\hat{k}) = 4 + 0 - 3 = 1 \] Thus, the equation becomes: \[ \vec{r} \cdot (2\hat{i} - 3\hat{j} + 3\hat{k}) - 1 = 0 \] **Final Equation of the Plane:** \[ \vec{r} \cdot (2\hat{i} - 3\hat{j} + 3\hat{k}) = 1 \] ---
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MODERN PUBLICATION-THREE DIMENSIONAL GEOMETRY -EXERCISE 11 (E) (SHORT ANSWER TYPE QUESTIONS )
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  7. Find the vector and cartesian equation of the plane : (i) that passe...

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  17. Find the angle between the lines x-2y+z=0=x+2y-2za n dx+2y+z=0=3x+9y+5...

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  18. Show that the line 3x - 2y + 5 = 0 , y + 3z - 15 = 0 and (x -1)/(5) =...

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  19. Find the equations of the line passing through the point (1, -2, 3) an...

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