Cross Product

Question

For the vectors a = i + j + k and b = 2i - j + 3k, determine the magnitude of their vector product a x b.

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The cross product of two vectors yields a third vector perpendicular to both, computed as the symbolic determinant with rows (i, j, k), the components of a, and the components of b. Its magnitude equals |a||b|sin(theta) and represents the area of the parallelogram spanned by the vectors, a recurring JEE Advanced theme. Expanding, a x b = i[(1)(3) - (1)(-1)] - j[(1)(3) - (1)(2)] + k[(1)(-1) - (1)(2)] = i(3 + 1) - j(3 - 2) + k(-1 - 2) = 4i - j - 3k. The magnitude is sqrt(4^2 + (-1)^2 + (-3)^2) = sqrt(16 + 1 + 9) = sqrt(26). Option sqrt(30) comes from mis-squaring the j-component as 5 instead of 1; option sqrt(27) drops a unit from the sum 16 + 1 + 9; option 6 reports |a||b| approximately without the sine factor. Plausibility check: the resulting vector dotted with a gives 4 - 1 - 3 = 0, confirming perpendicularity as required of any genuine cross product.

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Cross Product

Select the correct option:

🔒 Solution Hidden from View

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Solution

The magnitude of cross product of two vectors $a$ and $b$ is