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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2019 · Mathematics · Three Dimensional Geometry

The angle between the lines whose direction cosine satisfy the...

The angle between the lines whose direction cosine satisfy the equations l+m+n=0l + m + n = 0 and l2=m2+n2l^{2} = m^{2} + n^{2} is

Options

  1. Aπ3\dfrac{\pi}{3}
  2. Bπ4\dfrac{\pi}{4}
  3. Cπ6\dfrac{\pi}{6}
  4. Dπ2\dfrac{\pi}{2}

Correct answer

A. π3\dfrac{\pi}{3}

Step-by-step solution

We have,&

l+m+n=0l + m + n = 0 & ...(i). . . ( i )

and l2=m2+n2l^{2} = m^{2} + n^{2} & ...(ii). . . ( ii )

Putting the value of ll in Eq. (ii)( ii ), we get

(m+n)2=m2+n2( m + n{)}^{2} = m^{2} + n^{2}

⇒m2+n2+2mn=m2+n2⇒ m^{2} + n^{2} + 2mn = m^{2} + n^{2}

⇒mn=0⇒ mn = 0

⇒m=0,n=0⇒ m = 0, n = 0

when m=0,l=−nm = 0, l = - n

∴∴ Direction cosines are \left(- \dfrac{1}{2}, 0, \dfrac{1}{2}\right).</p><p>When.</p><p>Whenn = 0, l = - m</p><p></p><p>∴Directioncosinesare(−12,12,0)Direction cosines are \left(- \dfrac{1}{2}, \dfrac{1}{2}, 0\right).

Angle between lines are

cosθ=(−12×−12)+(0×12)+(12×0)cos\theta = \left(\dfrac{- 1}{2} \times - \dfrac{1}{2}\right) + \left(0 \times \dfrac{1}{2}\right) +\left(\dfrac{1}{2} \times0\right)

cos\theta = 12,θ=π\dfrac{1}{2}, \theta = \pi/3/ 3

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