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  • If the eqation of the plane containing the line <img alt="x+2 y+3 z-4=0\, 2 x+y-z+5" src="https://entrancecorner.oncodecogs.com/gif.latex?x&plus;2%20y&plus;3%20z-4%3D0%5C%2C%202%20x&plu

If the eqation of the plane containing the line x+2 y+3 z-4=0\, 2 x+y-z+5 and perpendicular to the plane \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) is  a x+b y+c z=4, then (a-b+c) is equal to

Option: 1

22


Option: 2

24


Option: 3

20


Option: 4

18


Answers (1)

best_answer

D.R’s of line \overrightarrow{\mathrm{n}}_1=-5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-3 \hat{\mathrm{k}}

D.R's of normal of second plane
$$ \begin{aligned} & \overrightarrow{\mathrm{n}}_2=5 \hat{i}-2 \hat{j}-3 \hat{k} \\ & \overrightarrow{\mathrm{n}}_1 \times \overrightarrow{\mathrm{n}}_2=-27 \hat{i}-30 \hat{j}-25 \hat{k} \end{aligned}
A point on the required plane is \left(0,-\frac{11}{5}, \frac{14}{5}\right)
The equation of required plane is
$$ \begin{aligned} & 27 \mathrm{x}+30 \mathrm{y}+25 \mathrm{z}=4 \\ & \therefore \mathrm{a}-\mathrm{b}+\mathrm{c}=22 \end{aligned}

Posted by

Irshad Anwar

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