A line 'l' passing through origin is perpendicular to the lines l_1:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat kl_2:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k If the co-ordinates of the point in the first octant on 'l2‘ at a distance of \sqrt 17 from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2021 · 25 Feb · Shift 2 · Q44 with Solution

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Correct answer: 44

Write the given lines in vector form

The lines are

\vec r=(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k$$ $$l_2: \vec r=(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k$$ So a point on $l_1$ is $$P_1(t)=(3+t,-1+2t,4+2t)$$ and its direction vector is $$\vec d_1=(1,2,2).$$ A point on $l_2$ is $$P_2(s)=(3+2s,3+2s,2+s)$$ and its direction vector is $$\vec d_2=(2,2,1).$$ --- 2. **Find the direction vector of line $l$** Since line $l$ passes through the origin and is perpendicular to both $l_1$ and $l_2$, its direction vector must be perpendicular to both $\vec d_1$ and $\vec d_2$. Hence it is parallel to $$\vec d_1\times \vec d_2.$$ Compute: $$\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&2&2\\ 2&2&1 \end{vmatrix}$$ $$=\hat i(2\cdot 1-2\cdot 2)-\hat j(1\cdot 1-2\cdot 2)+\hat k(1\cdot 2-2\cdot 2)$$ $$=(-2,3,-2).$$ So line $l$ is $$\vec r=\lambda(-2,3,-2).$$ --- 3. **Find the point of intersection of $l$ and $l_1$** Let the intersection point satisfy $$(-2\lambda,3\lambda,-2\lambda)=(3+t,-1+2t,4+2t).$$ Equating coordinates: $$-2\lambda=3+t \quad ...(1)$$ $$3\lambda=-1+2t \quad ...(2)$$ $$-2\lambda=4+2t \quad ...(3)$$ From (1): $$t=-2\lambda-3.$$ Substitute into (2): $$3\lambda=-1+2(-2\lambda-3)=-1-4\lambda-6=-7-4\lambda$$ $$7\lambda=-7$$ $$\lambda=-1.$$ Then $$t=-2(-1)-3=2-3=-1.$$ So the intersection point is $$P=(-2(-1),3(-1),-2(-1))=(2,-3,2).$$ --- 4. **Find the point on $l_2$ at distance $\sqrt{17}$ from $P$** A general point on $l_2$ is $$Q=(3+2s,3+2s,2+s).$$ We are given $$PQ=\sqrt{17}.$$ So $$[(3+2s)-2]^2+[(3+2s)-(-3)]^2+[(2+s)-2]^2=17.$$ That is, $$ (1+2s)^2+(6+2s)^2+s^2=17.$$ Expand: $$ (1+4s+4s^2)+(36+24s+4s^2)+s^2=17$$ $$9s^2+28s+37=17$$ $$9s^2+28s+20=0.$$ Solve: $$9s^2+28s+20=(9s+10)(s+2)=0.$$ Thus $$s=-\frac{10}{9} \quad \text{or} \quad s=-2.$$ --- 5. **Choose the point in the first octant** For $s=-2$: $$Q=(3-4,3-4,2-2)=(-1,-1,0),$$ not in the first octant. For $s=-\frac{10}{9}$: $$Q=\left(3-\frac{20}{9},3-\frac{20}{9},2-\frac{10}{9}\right) =\left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right).$$ This lies in the first octant. Hence $$a=\frac79,\quad b=\frac79,\quad c=\frac89.$$ --- 6. **Compute the required value** $$a+b+c=\frac79+\frac79+\frac89=\frac{22}{9}.$$ Therefore, $$18(a+b+c)=18\cdot \frac{22}{9}=2\cdot 22=44.$$ --- 7. **Comparison with stored answer** Derived answer = $44$. Stored correct answer = $44$. They agree.

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View written solutionFree

Correct answer: 44

Write the given lines in vector form

The lines are

\vec r=(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k$$ $$l_2: \vec r=(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k$$ So a point on $l_1$ is $$P_1(t)=(3+t,-1+2t,4+2t)$$ and its direction vector is $$\vec d_1=(1,2,2).$$ A point on $l_2$ is $$P_2(s)=(3+2s,3+2s,2+s)$$ and its direction vector is $$\vec d_2=(2,2,1).$$ --- 2. **Find the direction vector of line $l$** Since line $l$ passes through the origin and is perpendicular to both $l_1$ and $l_2$, its direction vector must be perpendicular to both $\vec d_1$ and $\vec d_2$. Hence it is parallel to $$\vec d_1\times \vec d_2.$$ Compute: $$\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&2&2\\ 2&2&1 \end{vmatrix}$$ $$=\hat i(2\cdot 1-2\cdot 2)-\hat j(1\cdot 1-2\cdot 2)+\hat k(1\cdot 2-2\cdot 2)$$ $$=(-2,3,-2).$$ So line $l$ is $$\vec r=\lambda(-2,3,-2).$$ --- 3. **Find the point of intersection of $l$ and $l_1$** Let the intersection point satisfy $$(-2\lambda,3\lambda,-2\lambda)=(3+t,-1+2t,4+2t).$$ Equating coordinates: $$-2\lambda=3+t \quad ...(1)$$ $$3\lambda=-1+2t \quad ...(2)$$ $$-2\lambda=4+2t \quad ...(3)$$ From (1): $$t=-2\lambda-3.$$ Substitute into (2): $$3\lambda=-1+2(-2\lambda-3)=-1-4\lambda-6=-7-4\lambda$$ $$7\lambda=-7$$ $$\lambda=-1.$$ Then $$t=-2(-1)-3=2-3=-1.$$ So the intersection point is $$P=(-2(-1),3(-1),-2(-1))=(2,-3,2).$$ --- 4. **Find the point on $l_2$ at distance $\sqrt{17}$ from $P$** A general point on $l_2$ is $$Q=(3+2s,3+2s,2+s).$$ We are given $$PQ=\sqrt{17}.$$ So $$[(3+2s)-2]^2+[(3+2s)-(-3)]^2+[(2+s)-2]^2=17.$$ That is, $$ (1+2s)^2+(6+2s)^2+s^2=17.$$ Expand: $$ (1+4s+4s^2)+(36+24s+4s^2)+s^2=17$$ $$9s^2+28s+37=17$$ $$9s^2+28s+20=0.$$ Solve: $$9s^2+28s+20=(9s+10)(s+2)=0.$$ Thus $$s=-\frac{10}{9} \quad \text{or} \quad s=-2.$$ --- 5. **Choose the point in the first octant** For $s=-2$: $$Q=(3-4,3-4,2-2)=(-1,-1,0),$$ not in the first octant. For $s=-\frac{10}{9}$: $$Q=\left(3-\frac{20}{9},3-\frac{20}{9},2-\frac{10}{9}\right) =\left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right).$$ This lies in the first octant. Hence $$a=\frac79,\quad b=\frac79,\quad c=\frac89.$$ --- 6. **Compute the required value** $$a+b+c=\frac79+\frac79+\frac89=\frac{22}{9}.$$ Therefore, $$18(a+b+c)=18\cdot \frac{22}{9}=2\cdot 22=44.$$ --- 7. **Comparison with stored answer** Derived answer = $44$. Stored correct answer = $44$. They agree.

3D Geometry question

3D Geometry question

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