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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)

best_answer

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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