CBSE 2024, Set 1

Question

**Assertion (A):** A line in space cannot be drawn perpendicular to x, y and z axes simultaneously.

**Reason (R):** For any line making angles $\alpha, \beta, \gamma$ with the positive directions of X, y and Z axes respectively, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. A). Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). B). Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). C). Assertion (A) is true, but Reason (R) is false. D). Assertion (A) is false, but Reason (R) is true.

Edvaya · Accepted Answer

The correct answer is Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).. **Solution:**

Let's test the hypothesis in Assertion (A) using the mathematical principle provided in Reason (R).

If a line is perpendicular to all three coordinate axes (x, y, and z) simultaneously, it means the direction angles it makes with the axes are all $90^\circ$:
$$\alpha = 90^\circ$$
$$\beta = 90^\circ$$
$$\gamma = 90^\circ$$

The direction cosines ($l, m, n$) would then be:
$$l = \cos(90^\circ) = 0$$
$$m = \cos(90^\circ) = 0$$
$$n = \cos(90^\circ) = 0$$

According to the fundamental property of direction cosines (stated in Reason R):
$$l^2 + m^2 + n^2 = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

Substituting our hypothetical values:
$$0^2 + 0^2 + 0^2 = 0$$

Since $0 
e 1$, this leads to a mathematical contradiction. A set of direction angles must satisfy the sum of squares equal to $1$. Because $\alpha=\beta=\gamma=90^\circ$ fails this condition, such a line cannot exist.

Therefore, Assertion (A) is True. Reason (R) is True and is the exact mathematical principle used to prove that the assertion is correct.

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Mathematics > Coordinate Geometry > Three Dimensional Geometry

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