Let the position vectors of two points P and Q be 3 \widehat i-\widehat j + 2 \widehat k and \widehat i + 2 \widehat j- 4 \widehat k , respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, - 1, 2) and ( - 2, 1, - 2), respectively. Let lines PR and QS intersect at T. If the vector \overrightarrow TA is perpendicular to both \overrightarrow PR and \overrightarrow QS and the length of vector \overrightarrow TA is \sqrt 5 units, then the modulus of a position vector of A is :

3D Geometry PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q25 with Solution

View written solutionFree

Correct answer: A

Write the given points and lines

The position vectors are $\vec{OP}=(3,-1,2), \qquad \vec{OQ}=(1,2,-4).$

The direction ratios are:

for line $PR$ : $(4,-1,2)$
for line $QS$ : $(-2,1,-2)$

So the parametric equations of the two lines are:

$PR: \vec r=(3,-1,2)+\lambda(4,-1,2)$

$QS: \vec r=(1,2,-4)+\mu(-2,1,-2)$

Thus a general point on $PR$ is

-1-\lambda, 2+2\lambda),$$ and a general point on $QS$ is $$T=(1-2\mu, 2+\mu, -4-2\mu).$$ Since the lines intersect at $T$, equate coordinates. --- 2. **Find the point of intersection $T$** From coordinates: $$3+4\lambda=1-2\mu \quad ...(1)$$ $$-1-\lambda=2+\mu \quad ...(2)$$ $$2+2\lambda=-4-2\mu \quad ...(3)$$ From (2): $$-\lambda-\mu=3 \implies \lambda+\mu=-3$$ From (3): $$2\lambda+2\mu=-6 \implies \lambda+\mu=-3$$ which is consistent. Using (1): $$4\lambda+2\mu=-2 \implies 2\lambda+\mu=-1$$ Subtracting $\lambda+\mu=-3$ from this: $$\lambda=2$$ Then $$\mu=-5$$ Hence $$T=(3+4\cdot2,-1-2,2+2\cdot2)=(11,-3,6).$$ --- 3. **Find a vector perpendicular to both given lines** Let $$\vec d_1=(4,-1,2), \qquad \vec d_2=(-2,1,-2).$$ A vector perpendicular to both is given by the cross product: $$\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & -1 & 2\\ -2 & 1 & -2 \end{vmatrix}$$ $$=\hat i\big(( -1)(-2)-2(1)\big)-\hat j\big(4(-2)-2(-2)\big)+\hat k\big(4(1)-(-1)(-2)\big)$$ $$=\hat i(2-2)-\hat j(-8+4)+\hat k(4-2)$$ $$=(0,4,2)=2(0,2,1).$$ So a simpler perpendicular direction is $$\vec n=(0,2,1).$$ Its magnitude is $$|\vec n|=\sqrt{0^2+2^2+1^2}=\sqrt5.$$ --- 4. **Use the condition $|\overrightarrow{TA}|=\sqrt5$** Since $\overrightarrow{TA}$ is perpendicular to both $\overrightarrow{PR}$ and $\overrightarrow{QS}$, it must be along $\vec n=(0,2,1)$ or its opposite. Also, its length is $\sqrt5$, same as $|\vec n|$. Hence $$\overrightarrow{TA}=\pm(0,2,1).$$ Therefore the position vector of $A$ is $$\vec{OA}=\vec{OT}+\vec{TA}=(11,-3,6)\pm(0,2,1).$$ So the two possible points are: $$A_1=(11,-1,7),$$ $$A_2=(11,-5,5).$$ --- 5. **Find the modulus of the position vector of $A$** For $A_1=(11,-1,7)$, $$|\vec{OA_1}|=\sqrt{11^2+(-1)^2+7^2}= \sqrt{121+1+49}= \sqrt{171}.$$ For $A_2=(11,-5,5)$, $$|\vec{OA_2}|=\sqrt{11^2+(-5)^2+5^2}= \sqrt{121+25+25}= \sqrt{171}.$$ So in either case, the modulus is $$\boxed{\sqrt{171}}.$$ --- 6. **Check options** - A: $\sqrt{171}$ ✅ - B: $\sqrt{227}$ ❌ - C: $\sqrt{482}$ ❌ - D: $\sqrt{5}$ ❌ Therefore, the correct option is **A**.

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View written solutionFree

Correct answer: A

Write the given points and lines

The position vectors are $\vec{OP}=(3,-1,2), \qquad \vec{OQ}=(1,2,-4).$

The direction ratios are:

for line $PR$ : $(4,-1,2)$
for line $QS$ : $(-2,1,-2)$

So the parametric equations of the two lines are:

$PR: \vec r=(3,-1,2)+\lambda(4,-1,2)$

$QS: \vec r=(1,2,-4)+\mu(-2,1,-2)$

Thus a general point on $PR$ is

-1-\lambda, 2+2\lambda),$$ and a general point on $QS$ is $$T=(1-2\mu, 2+\mu, -4-2\mu).$$ Since the lines intersect at $T$, equate coordinates. --- 2. **Find the point of intersection $T$** From coordinates: $$3+4\lambda=1-2\mu \quad ...(1)$$ $$-1-\lambda=2+\mu \quad ...(2)$$ $$2+2\lambda=-4-2\mu \quad ...(3)$$ From (2): $$-\lambda-\mu=3 \implies \lambda+\mu=-3$$ From (3): $$2\lambda+2\mu=-6 \implies \lambda+\mu=-3$$ which is consistent. Using (1): $$4\lambda+2\mu=-2 \implies 2\lambda+\mu=-1$$ Subtracting $\lambda+\mu=-3$ from this: $$\lambda=2$$ Then $$\mu=-5$$ Hence $$T=(3+4\cdot2,-1-2,2+2\cdot2)=(11,-3,6).$$ --- 3. **Find a vector perpendicular to both given lines** Let $$\vec d_1=(4,-1,2), \qquad \vec d_2=(-2,1,-2).$$ A vector perpendicular to both is given by the cross product: $$\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & -1 & 2\\ -2 & 1 & -2 \end{vmatrix}$$ $$=\hat i\big(( -1)(-2)-2(1)\big)-\hat j\big(4(-2)-2(-2)\big)+\hat k\big(4(1)-(-1)(-2)\big)$$ $$=\hat i(2-2)-\hat j(-8+4)+\hat k(4-2)$$ $$=(0,4,2)=2(0,2,1).$$ So a simpler perpendicular direction is $$\vec n=(0,2,1).$$ Its magnitude is $$|\vec n|=\sqrt{0^2+2^2+1^2}=\sqrt5.$$ --- 4. **Use the condition $|\overrightarrow{TA}|=\sqrt5$** Since $\overrightarrow{TA}$ is perpendicular to both $\overrightarrow{PR}$ and $\overrightarrow{QS}$, it must be along $\vec n=(0,2,1)$ or its opposite. Also, its length is $\sqrt5$, same as $|\vec n|$. Hence $$\overrightarrow{TA}=\pm(0,2,1).$$ Therefore the position vector of $A$ is $$\vec{OA}=\vec{OT}+\vec{TA}=(11,-3,6)\pm(0,2,1).$$ So the two possible points are: $$A_1=(11,-1,7),$$ $$A_2=(11,-5,5).$$ --- 5. **Find the modulus of the position vector of $A$** For $A_1=(11,-1,7)$, $$|\vec{OA_1}|=\sqrt{11^2+(-1)^2+7^2}= \sqrt{121+1+49}= \sqrt{171}.$$ For $A_2=(11,-5,5)$, $$|\vec{OA_2}|=\sqrt{11^2+(-5)^2+5^2}= \sqrt{121+25+25}= \sqrt{171}.$$ So in either case, the modulus is $$\boxed{\sqrt{171}}.$$ --- 6. **Check options** - A: $\sqrt{171}$ ✅ - B: $\sqrt{227}$ ❌ - C: $\sqrt{482}$ ❌ - D: $\sqrt{5}$ ❌ Therefore, the correct option is **A**.

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