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Consider a line r⃗=(i^+4j^)+λ(2i^−2j^+3k^)\vec{r} = (\hat{i} + 4\hat{j}) + \lambda(2\hat{i} - 2\hat{j} + 3\hat{k}), then which of the following statements are correct?

(A) it passes through point (9, -4, 12)

(B) it passes through point (1, 4, -1)

(C) its direction cosine's are 217,−217,317\frac{2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}

(D) its Cartesian equation is x−12=y−4−2=z3\frac{x - 1}{2} = \frac{y - 4}{-2} = \frac{z}{3}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

Given line: r⃗=(i^+4j^)+λ(2i^−2j^+3k^)\vec{r} = (\hat{i} + 4\hat{j}) + \lambda(2\hat{i} - 2\hat{j} + 3\hat{k})

The line can be written as:

Point on line: (1,4,0)(1, 4, 0)

Direction vector: (2,−2,3)(2, -2, 3)


For point (9,−4,12)(9, -4, 12):

(x,y,z)=(1,4,0)+λ(2,−2,3)(x, y, z) = (1, 4, 0) + \lambda(2, -2, 3)

(x,y,z)=(1+2λ,4−2λ,3λ)(x, y, z) = (1+2\lambda, 4-2\lambda, 3\lambda)

From x-coordinate:

1+2λ=91 + 2\lambda = 9

λ=4\lambda = 4

From y-coordinate:

4−2λ=−44 - 2\lambda = -4

λ=4\lambda = 4

From z-coordinate:

3λ=123\lambda = 12

λ=4\lambda = 4

All three coordinates give λ=4\lambda = 4.

Statement (A) is correct.


For point (1,4,−1)(1, 4, -1):

From x-coordinate:

1+2λ=11 + 2\lambda = 1

λ=0\lambda = 0

From y-coordinate:

4−2λ=44 - 2\lambda = 4

λ=0\lambda = 0

From z-coordinate:

3λ=−13\lambda = -1

λ=−13\lambda = -\frac{1}{3}

Different values of λ\lambda are obtained.

Statement (B) is incorrect.


Direction vector: (2,−2,3)(2, -2, 3)

Magnitude:

∣d⃗∣=22+(−2)2+32|\vec{d}| = \sqrt{2^2 + (-2)^2 + 3^2}

∣d⃗∣=4+4+9|\vec{d}| = \sqrt{4 + 4 + 9}

∣d⃗∣=17|\vec{d}| = \sqrt{17}

Direction cosines:

(217,−217,317)\left(\frac{2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right)

Statement (C) is correct.


For a line through point (x0,y0,z0)(x_0, y_0, z_0) with direction vector (a,b,c)(a, b, c):

x−x0a=y−y0b=z−z0c\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}

Using point (1,4,0)(1, 4, 0) and direction (2,−2,3)(2, -2, 3):

x−12=y−4−2=z−03\frac{x - 1}{2} = \frac{y - 4}{-2} = \frac{z - 0}{3}

x−12=y−4−2=z3\frac{x - 1}{2} = \frac{y - 4}{-2} = \frac{z}{3}

Statement (D) is correct.


Correct statements: (A), (C), and (D)

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