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MHTCET 10 May 2024 Shift 2 Paper

Question : 133 of 150

Marks: +1, -0

\overline{\mathrm{AB}}

\overline{\mathrm{CD}}

A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)

D \equiv(7,0,10)

Solution:

Concept:

The projection of one vector on another is given by the dot product divided by the magnitude of the second vector.

Explanation:

\overline{\mathrm{AB}} = B - A = (1-2,\,-4+3,\,-2-0) = (-1,\,-1,\,-2)

So,

\overline{\mathrm{AB}} = -\hat{i} - \hat{j} - 2\hat{k}

\overline{\mathrm{CD}} = D - C = (7-4,\,0-6,\,10-8) = (3,\,-6,\,2)

So,

\overline{\mathrm{CD}} = 3\hat{i} - 6\hat{j} + 2\hat{k}

Scalar projection of

\overline{\mathrm{AB}}

\overline{\mathrm{CD}}

is:

\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}}{|\overline{\mathrm{CD}}|}

\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}} = (-1)(3) + (-1)(-6) + (-2)(2) = -3 + 6 - 4 = -1

|\overline{\mathrm{CD}}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = 7

Projection

= \frac{-1}{7}

. The magnitude of this projection is

\frac{1}{7}

Answer:

\frac{1}{7}

, i.e. Option B.

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