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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Probability

BITSAT 2010 · Mathematics · Probability

The equation of the plane containing the line ( x-x_1 = y-y_1 ~m =...

The equation of the plane containing the line (x−x1ℓ=y−y1 m=z−z1n\frac{\mathrm{x}-\mathrm{x}_1}{\ell}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{n}}) is (a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0), then

Options

  1. Aax1+by1+cz1=0)\mathrm{ax}_{1}+ \mathrm{by}_{1}+ \mathrm{cz}_{1}=0)
  2. B(a \ell+b m+c n=0)
  3. C(aℓ=bm=cn\frac{\mathrm{a}}{\ell}=\frac{\mathrm{b}}{\mathrm{m}}=\frac{\mathrm{c}}{\mathrm{n}})
  4. D(\ell x1+my1+nz1=\mathrm{x}_{1}+ \mathrm{my}_{1}+ \mathrm{nz}_{1}=0)

Correct answer

B. (a \ell+b m+c n=0)

Step-by-step solution

If the given plane contains the given line then the normal to the plane, must be perpendicular to the line and the condition for the same is (a \ell+b m+c n=0).
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