Home/Class 12/Maths/

## QuestionMathsClass 12

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) $$\dfrac{1}{{\sqrt 2 }}$$

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 $$-\dfrac{1}{\sqrt{3}}$$