Question:
If the equation of the plane containing the line x+2y+3z-4=0=2x+y-z+5 and perpendicular to the plane \(\vec{r}=(\vec{i}-\vec{j})+\lambda(\vec{i}+\vec{j}+\vec{k})+\mu(\vec{i}-2\vec{j}+3\vec{k})\) is ax+by+cz=4, then (a-b+c) is equal to
If the equation of the plane containing the line x+2y+3z-4=0=2x+y-z+5 and perpendicular to the plane \(\vec{r}=(\vec{i}-\vec{j})+\lambda(\vec{i}+\vec{j}+\vec{k})+\mu(\vec{i}-2\vec{j}+3\vec{k})\) is ax+by+cz=4, then (a-b+c) is equal to
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To find the equation of a plane, use points and direction ratios from the cross-product of vectors.
Updated On: Mar 11, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Find the direction ratios (D.R.s).
- The D.R.s of the line are obtained by taking the cross-product of the given vectors: \[ \vec{r}_1 = \hat{i} - \hat{j}, \quad \vec{r}_2 = \hat{i} + \hat{j} + \hat{k}, \quad \vec{r}_3 = \hat{i} - 2\hat{j} + 3\hat{k}. \] \[ \vec{r}_1 \times \vec{r}_2 = (-27\hat{i} - 30\hat{j} - 25\hat{k}). \] Step 2: Find the equation of the plane.
- A point on the plane is \((0, \frac{11}{5}, \frac{14}{5})\). - Substituting into the general equation: \[ 27x + 30y + 25z = 4. \] Step 3: Calculate \(a - b + c\).
- From the equation, \(a = 27\), \(b = 30\), \(c = 25\). \[ a - b + c = 27 - 30 + 25 = 22. \] Final Answer: \(a - b + c = 22\).
- The D.R.s of the line are obtained by taking the cross-product of the given vectors: \[ \vec{r}_1 = \hat{i} - \hat{j}, \quad \vec{r}_2 = \hat{i} + \hat{j} + \hat{k}, \quad \vec{r}_3 = \hat{i} - 2\hat{j} + 3\hat{k}. \] \[ \vec{r}_1 \times \vec{r}_2 = (-27\hat{i} - 30\hat{j} - 25\hat{k}). \] Step 2: Find the equation of the plane.
- A point on the plane is \((0, \frac{11}{5}, \frac{14}{5})\). - Substituting into the general equation: \[ 27x + 30y + 25z = 4. \] Step 3: Calculate \(a - b + c\).
- From the equation, \(a = 27\), \(b = 30\), \(c = 25\). \[ a - b + c = 27 - 30 + 25 = 22. \] Final Answer: \(a - b + c = 22\).
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