The point of the intersection of the straight lines {r} = (3{i} - 4{j} + 5{k}) + λ(-{i} - 2{j} + 2{k}) and 3-x/-1 = y+4/2 = z-5/7 is

Correct Answer is : (3, – 4, 5)

Correct Answer is : (3, – 4, 5)

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KEAM 2016 Math Paper

Question : 78 of 120

Marks: +1, -0

\vec{r} = (3\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(-\hat{i} - 2\hat{j} + 2\hat{k})

\frac{3-x}{-1} = \frac{y+4}{2} = \frac{z-5}{7}

(– 3, – 4, – 5)
(– 3, 4, 5)
(– 3, 4, – 5)
(– 3, – 4, 5)
(3, – 4, 5)

Solution:

Given equation of straight lines are

r = (3\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(-\hat{i} - 2\hat{j} + 2\hat{k})

\frac{x-3}{-1} = \frac{y+4}{2} = \frac{z-5}{2} = \lambda

......(i)

and

\frac{3-x}{-1} = \frac{y+4}{2} = \frac{z-5}{7} = \mu

or,

\frac{x-3}{1} = \frac{y+4}{2} = \frac{z-5}{7} = \mu

.....(ii)

From Eq. (i) , coordinates are

(-\lambda+3-2\lambda-4, 2\lambda+5)

From Eq. (ii) coordinates are

(\mu+3, 2\mu-4, 7\mu+5)

For intersecting the given lines,

-\lambda+3 = \mu+3

\Rightarrow \mu + \lambda = 0

.....(iii)

-2\lambda-4 = 2\mu-4

\Rightarrow 2\mu + 2\lambda = 0

....(iv)

2\lambda+5 = 7\mu+5

\Rightarrow 7\mu - 2\lambda = 0

....(v)

From Eqs. (iii), (vi), (v)

\mu=0, \lambda=0

Hence, intersecting points are

(0+3, -2 \times 0 -4, 2 \times 0 +5) = (3, -4, 5)

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