The projections of the line segment joining points P(-1, 2, 3) and Q(1, 0, 4) on a line inclined equal acute angles with the coordinates axis is

(A) \(-\dfrac{1}{{\sqrt 3 }}\)

(B) \(\sqrt 3 \)

(C) \(\dfrac{1}{3}\)

(D) \(\dfrac{1}{{\sqrt 2 }}\)

(A) \(-\dfrac{1}{{\sqrt 3 }}\)

(B) \(\sqrt 3 \)

(C) \(\dfrac{1}{3}\)

(D) \(\dfrac{1}{{\sqrt 2 }}\)

Answer: A

Consider the points be \(P\left(-1, 2, 3\right)\) and \(Q\left(1, 0, 4\right)\)

Now,\({\cos}^{2}{\alpha}+{\cos}^{2}{\beta}+{\cos}^{2}{\gamma}=1\)

\(3{\cos}^{2}{\alpha}=1\) since \(\alpha=\beta=\gamma\)

\(\Rightarrow\,\cos{\alpha}=\dfrac{1}{\sqrt{3}}\)

Then,\(\vec{v}=L\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)\)

\(\vec{u}=\vec{PQ}=\vec{OQ}-\vec{OP}\)

\(=\left(\hat{i}+4\hat{k}\right)-\left(-\hat{i}+2\hat{j}+3\hat{k}\right)=-2\hat{j}+\hat{k}\)

\(\vec{u}=\left(0,-2,1\right)\)

Now,\(\left|\vec{u}\right|\cos{\theta}=\left|\vec{u}\right|\dfrac{\left(\vec{u}.\vec{v}\right)}{\left|\vec{u}\right|\left|\vec{v}\right|}\)

\(=\dfrac{0\times\dfrac{1}{\sqrt{3}}-2\times\dfrac{1}{\sqrt{3}}+1\times\dfrac{1}{\sqrt{3}}}{\sqrt{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}}\)

\(=\dfrac{-\dfrac{2}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}}{\sqrt{\dfrac{3}{3}}}\)

\(=-\dfrac{1}{\sqrt{3}}\)

Hence the projection is \(-\dfrac{1}{\sqrt{3}}\)

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