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Find the vector and cartesian equation of a plane which passes through the point `(2,-1,3)` and perpendicular to a line whose d.r.'s are `1,-3,5`.

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To find the vector and Cartesian equation of a plane that passes through the point \( (2, -1, 3) \) and is perpendicular to a line with direction ratios \( 1, -3, 5 \), we can follow these steps: ### Step 1: Identify the Point and Direction Ratios We have a point \( A(2, -1, 3) \) through which the plane passes. The direction ratios of the line are given as \( 1, -3, 5 \). ### Step 2: Determine the Normal Vector Since the plane is perpendicular to the line, the direction ratios of the line will serve as the normal vector \( \mathbf{n} \) of the plane. Thus, we can write: \[ \mathbf{n} = \langle 1, -3, 5 \rangle \] ### Step 3: Write the Vector Equation of the Plane The vector equation of a plane can be expressed as: \[ \mathbf{r} - \mathbf{A} \cdot \mathbf{n} = 0 \] where \( \mathbf{A} \) is the position vector of the point \( A \) and \( \mathbf{r} \) is the position vector of any point on the plane. The position vector \( \mathbf{A} \) is: \[ \mathbf{A} = 2\mathbf{i} - 1\mathbf{j} + 3\mathbf{k} \] Substituting \( \mathbf{A} \) and \( \mathbf{n} \) into the equation gives: \[ \mathbf{r} \cdot \langle 1, -3, 5 \rangle - (2\mathbf{i} - 1\mathbf{j} + 3\mathbf{k}) \cdot \langle 1, -3, 5 \rangle = 0 \] ### Step 4: Calculate \( \mathbf{A} \cdot \mathbf{n} \) Now we need to calculate the dot product \( \mathbf{A} \cdot \mathbf{n} \): \[ \mathbf{A} \cdot \mathbf{n} = (2)(1) + (-1)(-3) + (3)(5) = 2 + 3 + 15 = 20 \] ### Step 5: Write the Final Vector Equation Now we can write the vector equation of the plane: \[ \mathbf{r} \cdot \langle 1, -3, 5 \rangle = 20 \] ### Step 6: Convert to Cartesian Equation Let \( \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \). The dot product gives: \[ x(1) + y(-3) + z(5) = 20 \] This simplifies to: \[ x - 3y + 5z = 20 \] ### Final Answers - **Vector Equation of the Plane**: \(\mathbf{r} \cdot \langle 1, -3, 5 \rangle = 20\) - **Cartesian Equation of the Plane**: \(x - 3y + 5z = 20\)
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NAGEEN PRAKASHAN ENGLISH-THREE-DIMENSIONAL GEOMETRY -Exercise 11 D
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