(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
