# A butterfly is moving in a straight path in the space. Let this path be denoted by a line l, whose equation is (x−1)/ 2 = (y-2)/3 = (z −3)/ 4 (say).

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A butterfly is moving in a straight path in the space. Let this path be denoted by a line $l$, whose equation is $\cfrac{x-1}2=\cfrac{y-2}3=\cfrac{z-3}4$ (say). Using this information answer the following questions :

(i). What are the direction ratios and cosines of the line ?

(ii). Write the vector equation of the given line $l$.

(iii). If the z - coordinate of a point on this line $l$ is 11, then what is the x - coordinate of the same point on this line $l$?

(iv). The possible position vector of a point on the line $l$ is (v). Find the unit vector in the direction of the vector parallel to the given line $l$.

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(i). Given that the equation of line  is $\cfrac{x-1}2=\cfrac{y-2}3=\cfrac{z-3}4$.

The direction ratios of the line  are 2, 3, 4.

The direction cosines of the line  are Hence, the direction ratios and direction cosines of the line  are 2, 3, 4 and $\cfrac2{\sqrt{29}},\cfrac3{\sqrt{29}}$,$\cfrac4{\sqrt{29}}$, respectively.

(ii). Given that the equation of line  is $\cfrac{x-1}2=\cfrac{y-2}3=\cfrac{z-3}4$.

The direction ratios of the line  are 2, 3, 4 and line  is passing through the point (1, 2, 3).

The position vector of point (1, 2, 3) is $\vec a=\hat i+2\hat j+3\hat k.$

And the vector $\vec b=2\hat i+3\hat j+4\hat k.$.

Therefore, the line  is passing through the point (1, 2, 3) and parallel to the vector  $\vec b$.

The vector equation of line  is $\vec r=\vec a+\lambda\vec b$

$\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$

Hence, the vector equation of line  is $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$

(iii). Given that the equation of line  is $\cfrac{x-1}2=\cfrac{y-2}3=\cfrac{z-3}4$ = λ(Let)

⇒ x = 2λ + 1, y = 3λ + 2 and z = 4λ + 3.

Therefore, the points on the line  is of the form (2λ + 1, 3λ + 2, 4λ + 3).

Given that the z -coordinate of a point on this line  is 11.

Therefore, 4λ + 3 = 11

⇒ 4λ = 11 − 3 = 8

⇒ λ = 2.

Now, the x - coordinate of the same point on this line  is x = 2λ + 1 = 2 × 2 + 1 = 5.

(iv). Since, the points on the line  is of the form (2λ + 1, 3λ + 2, 4λ + 3).

Therefore, the position vector of the points on the line  is $\vec a$λ = (1 + 2λ)$\hat i$+ (2 + 3λ)$\hat j$+ (3 + 4λ)$\hat k$.

If λ = 0, then the position vector of the point on the line  is $\vec a_0$ = $\hat i+2\hat j+3\hat k.$.

If λ = $\cfrac12$, then the position vector of the point on the line  is $\vec a_{\frac12}$= $2\hat i+\cfrac72\hat j+5\hat k$.

Therefore, the possible position vector of the point on the line  among in the given options is$\hat i+2\hat j+3\hat k$.

Hence, option (a) is correct.

(v). The direction ratios of the line  : $\cfrac{x-1}2=\cfrac{y-2}3=\cfrac{z-3}4$ are 2, 3, 4.

Therefore, the vector parallel to the line  is $\vec b=2\hat i+3\hat j+4\hat k$.

(Because, we know that the line is parallel to the vector obtaining by the direction ratios of the line)

The magnitude of the vector $\vec b$ is $\sqrt{2^2+3^2+4^2}=\sqrt{4+9+16}$ = $\sqrt{29}$.

The unit vector in the direction of the vector parallel to the given line  is Hence, the unit vector in the direction of the vector parallel to the given line  is 