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NCERT Exemplar · Q19

Q.State whether the following statement is True or False: The vector equation of the line x−53=y+47=z−62\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2} is r⃗=5i^−4j^+6k^+λ(3i^+7j^+2k^)\vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k}).

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The statement is True. The given Cartesian equation directly gives the point (5,−4,6)(5,-4,6) and direction ratios (3,7,2)(3,7,2), which match the vector form exactly.


The core idea here is the translation between Cartesian and vector forms of a line. Every line in 3D can be written as:

r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}

where a⃗\vec{a} is the position vector of a fixed point on the line, and b⃗\vec{b} is a vector parallel to the line (the direction vector). The Cartesian form x−x1a=y−y1b=z−z1c\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} is just another way of saying the same thing — the denominators are the direction ratios, and (x1,y1,z1)(x_1, y_1, z_1) is a point on the line.

So the question is simply: does the given vector equation correctly extract these two pieces from the Cartesian equation?


  1. Identify the fixed point from the Cartesian form.

    The given line is x−53=y+47=z−62\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}.

    In the standard form x−x1a\dfrac{x-x_1}{a}, the numerator is (x−x1)(x - x_1). Here x−5x-5 means x1=5x_1 = 5.

    For yy, we have y+4y+4 which is y−(−4)y - (-4), so y1=−4y_1 = -4.

    For zz, z−6z-6 gives z1=6z_1 = 6.

    So the point is (5,−4,6)(5, -4, 6), whose position vector is 5i^−4j^+6k^5\hat{i} - 4\hat{j} + 6\hat{k}.

    This matches the a⃗\vec{a} in the given vector equation.

  2. Identify the direction vector.

    The denominators are 3,7,23, 7, 2 — these are the direction ratios.

    So a direction vector parallel to the line is 3i^+7j^+2k^3\hat{i} + 7\hat{j} + 2\hat{k}. …

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