A line with direction cosines proportional to 2, 1, 2 meets each of the lines x=y+a=z  and x+a=2y=2z. The co­ordinates of each of the points of intersection are given by

  • Option 1)

    (3a,2a,3a),(a,a,2a)

  • Option 2)

    (3a,2a,3a),(a,a,a)

  • Option 3)

    (3a,3a,3a),(a,a,a)

  • Option 4)

    (2a,3a,3a),(2a,a,a)

 

Answers (1)

As we learnt in 

Direction Cosines -

i)    \sin^{2} \alpha+ \sin^{2} \beta+\sin^{2} \gamma= 2

ii)    If OP =r then the co-ordinates of P will be (lr,mr,nr)

iii)    Direction cosines of X-axis are (1,0,0)

iv)    Direction cosines of Y-axis are (0,1,0)

v)    Direction cosines of Z-axis are (0,0,1)

-

 

 

Cartesian eqution of a line -

The equation of a line passing through two points A\left ( x_{0},y_{0},z_{0} \right )and parallel to vector having direction ratios as \left ( a,b,c \right )is given by

\frac{x-x_{0}}{a}= \frac{y-y_{0}}{b}= \frac{z-z_{0}}{c}

The equation of a line passing through two points A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right ) is given by

\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

- wherein

 

 L_{1}:\frac{x}{1}= \frac{y+a}{1}= \frac{z}{1}=k

 

Point is \left ( k,k-a,k \right )\, \, and \, \,point\, \, on \, \,L_{2}\, \,is \left ( 2l-a,l,l \right )\, \,where\, \, k\, \,and \, \,l\, \, are\, \,same\, \,constants.

Now DCs are proportional to (2, 1, 2)

 

So \frac{2l-a-k}{2}=\frac{l-k+a}{1}=\frac{l-k}{2}

On solving k=3a,\ l=a

We get points (3a,2a,3a) and (a,a,a).

 


Option 1)

(3a,2a,3a),(a,a,2a)

Incorrect Option

 

Option 2)

(3a,2a,3a),(a,a,a)

Correct Option

 

Option 3)

(3a,3a,3a),(a,a,a)

Incorrect Option

 

Option 4)

(2a,3a,3a),(2a,a,a)

Incorrect Option

 

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