The unit vector perpendicular to line L (DRs: 1, -3, 2 ) and the normal to the plane P_1 : x + y + z - 6 = 0 is: - Mathematics MCQ Mathematics Mock Test - 11 2026 | ExamTest.live

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1 mark

The unit vector perpendicular to line L (DRs: 1, - 3, 2) and the normal to the plane P_{1} : x + y + z - 6 = 0 is:

Solution & Step-by-step Explanation

Vector along line L : b = \hat{i} - 3 \hat{j} + 2 \hat{k} Vector normal to plane P_{1} : a = \hat{i} + \hat{j} + \hat{k} A vector perpendicular to both is a \times b : a \times b = \hat{i} 11 \hat{j} 1 - 3 \hat{k} 12 = \hat{i} (2 - (- 3)) - \hat{j} (2 - 1) + \hat{k} (- 3 - 1) = 5 \hat{i} - \hat{j} - 4 \hat{k} Magnitude ∣ a \times b ∣ = 5^{2} + (- 1)^{2} + (- 4)^{2} = 25 + 1 + 16 = 42 .The unit vector is \pm \frac{5 i ^ - j ^ - 4 k ^}{42} .

Try it yourself before checking the explanation above.

The unit vector perpendicular to line L (DRs: 1, - 3, 2) and the normal to the plane P_{1} : x + y + z - 6 = 0 is:

\pm \frac{( 5 i ^ + j ^ - 4 k ^ )}{42}

\pm \frac{( 5 i ^ - j ^ + 4 k ^ )}{42}

\pm \frac{( 5 i ^ - j ^ - 4 k ^ )}{42}

\pm \frac{( 5 i ^ + j ^ + 4 k ^ )}{42}

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