If the vectors \overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k , \overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k and \overrightarrow c = 2\widehat i + 3\widehat j + \widehat k are coplanar and the projection of \overrightarrow a on the vector \overrightarrow b is \sqrt 54 units, then the sum of all possible values of \lambda + \mu is equal to :

Vector Algebra PYQ — JEE Main 2023 · 29 Jan · Shift 1 · Q39 with Solution

View written solutionFree

Correct answer: A

Given vectors

$\vec a=(\lambda,\mu,4),\qquad \vec b=(-2,4,-2),\qquad \vec c=(2,3,1)$

We are told:

$\vec a,\vec b,\vec c$ are coplanar
projection of $\vec a$ on $\vec b$ is $\sqrt{54}$ units

We need the sum of all possible values of $\lambda+\mu$ .

Condition for coplanarity

Three vectors are coplanar if their scalar triple product is zero:

$[\vec a\ \vec b\ \vec c]=0$

So,

\lambda & \mu & 4\\ -2 & 4 & -2\\ 2 & 3 & 1 \end{vmatrix}=0$$ Expanding along the first row: $$\lambda\begin{vmatrix}4 & -2\\3 & 1\end{vmatrix}-\mu\begin{vmatrix}-2 & -2\\2 & 1\end{vmatrix}+4\begin{vmatrix}-2 & 4\\2 & 3\end{vmatrix}=0$$ Now compute each determinant: $$\begin{vmatrix}4 & -2\\3 & 1\end{vmatrix}=4\cdot1-(-2)\cdot3=4+6=10$$ $$\begin{vmatrix}-2 & -2\\2 & 1\end{vmatrix}=(-2)(1)-(-2)(2)=-2+4=2$$ $$\begin{vmatrix}-2 & 4\\2 & 3\end{vmatrix}=(-2)(3)-4(2)=-6-8=-14$$ Hence, $$10\lambda-2\mu+4(-14)=0$$ $$10\lambda-2\mu-56=0$$ $$5\lambda-\mu=28$$ So, $$\mu=5\lambda-28 \qquad ...(1)$$ --- 3. **Projection condition** Projection of $\vec a$ on $\vec b$ has magnitude $$\left|\operatorname{proj}_{\vec b}\vec a\right|=\frac{|\vec a\cdot\vec b|}{|\vec b|}$$ Given this equals $\sqrt{54}=3\sqrt6$. First compute: $$\vec a\cdot\vec b=\lambda(-2)+\mu(4)+4(-2)=-2\lambda+4\mu-8$$ Also, $$|\vec b|=\sqrt{(-2)^2+4^2+(-2)^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6$$ Thus, $$\frac{| -2\lambda+4\mu-8 |}{2\sqrt6}=3\sqrt6$$ Multiply both sides by $2\sqrt6$: $$|-2\lambda+4\mu-8|=3\sqrt6\cdot 2\sqrt6=36$$ So, $$-2\lambda+4\mu-8=\pm 36$$ --- 4. **Substitute the coplanarity relation** Using $\mu=5\lambda-28$: $$-2\lambda+4(5\lambda-28)-8=\pm36$$ $$-2\lambda+20\lambda-112-8=\pm36$$ $$18\lambda-120=\pm36$$ So two cases arise. ### Case 1: $$18\lambda-120=36$$ $$18\lambda=156$$ $$\lambda=\frac{26}{3}$$ Then $$\mu=5\cdot\frac{26}{3}-28=\frac{130}{3}-\frac{84}{3}=\frac{46}{3}$$ Hence, $$\lambda+\mu=\frac{26}{3}+\frac{46}{3}=\frac{72}{3}=24$$ ### Case 2: $$18\lambda-120=-36$$ $$18\lambda=84$$ $$\lambda=\frac{14}{3}$$ Then $$\mu=5\cdot\frac{14}{3}-28=\frac{70}{3}-\frac{84}{3}=-\frac{14}{3}$$ Hence, $$\lambda+\mu=\frac{14}{3}-\frac{14}{3}=0$$ --- 5. **Sum of all possible values of $\lambda+\mu$** Possible values are: $$24,\ 0$$ Their sum is $$24+0=24$$ --- 6. **Evaluate options** - **A: 24** ✅ - **B: 0** ❌ - **C: 18** ❌ - **D: 6** ❌ Therefore, the correct option is **A**.

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View written solutionFree

Correct answer: A

Given vectors

$\vec a=(\lambda,\mu,4),\qquad \vec b=(-2,4,-2),\qquad \vec c=(2,3,1)$

We are told:

$\vec a,\vec b,\vec c$ are coplanar
projection of $\vec a$ on $\vec b$ is $\sqrt{54}$ units

We need the sum of all possible values of $\lambda+\mu$ .

Condition for coplanarity

Three vectors are coplanar if their scalar triple product is zero:

$[\vec a\ \vec b\ \vec c]=0$

So,

\lambda & \mu & 4\\ -2 & 4 & -2\\ 2 & 3 & 1 \end{vmatrix}=0$$ Expanding along the first row: $$\lambda\begin{vmatrix}4 & -2\\3 & 1\end{vmatrix}-\mu\begin{vmatrix}-2 & -2\\2 & 1\end{vmatrix}+4\begin{vmatrix}-2 & 4\\2 & 3\end{vmatrix}=0$$ Now compute each determinant: $$\begin{vmatrix}4 & -2\\3 & 1\end{vmatrix}=4\cdot1-(-2)\cdot3=4+6=10$$ $$\begin{vmatrix}-2 & -2\\2 & 1\end{vmatrix}=(-2)(1)-(-2)(2)=-2+4=2$$ $$\begin{vmatrix}-2 & 4\\2 & 3\end{vmatrix}=(-2)(3)-4(2)=-6-8=-14$$ Hence, $$10\lambda-2\mu+4(-14)=0$$ $$10\lambda-2\mu-56=0$$ $$5\lambda-\mu=28$$ So, $$\mu=5\lambda-28 \qquad ...(1)$$ --- 3. **Projection condition** Projection of $\vec a$ on $\vec b$ has magnitude $$\left|\operatorname{proj}_{\vec b}\vec a\right|=\frac{|\vec a\cdot\vec b|}{|\vec b|}$$ Given this equals $\sqrt{54}=3\sqrt6$. First compute: $$\vec a\cdot\vec b=\lambda(-2)+\mu(4)+4(-2)=-2\lambda+4\mu-8$$ Also, $$|\vec b|=\sqrt{(-2)^2+4^2+(-2)^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6$$ Thus, $$\frac{| -2\lambda+4\mu-8 |}{2\sqrt6}=3\sqrt6$$ Multiply both sides by $2\sqrt6$: $$|-2\lambda+4\mu-8|=3\sqrt6\cdot 2\sqrt6=36$$ So, $$-2\lambda+4\mu-8=\pm 36$$ --- 4. **Substitute the coplanarity relation** Using $\mu=5\lambda-28$: $$-2\lambda+4(5\lambda-28)-8=\pm36$$ $$-2\lambda+20\lambda-112-8=\pm36$$ $$18\lambda-120=\pm36$$ So two cases arise. ### Case 1: $$18\lambda-120=36$$ $$18\lambda=156$$ $$\lambda=\frac{26}{3}$$ Then $$\mu=5\cdot\frac{26}{3}-28=\frac{130}{3}-\frac{84}{3}=\frac{46}{3}$$ Hence, $$\lambda+\mu=\frac{26}{3}+\frac{46}{3}=\frac{72}{3}=24$$ ### Case 2: $$18\lambda-120=-36$$ $$18\lambda=84$$ $$\lambda=\frac{14}{3}$$ Then $$\mu=5\cdot\frac{14}{3}-28=\frac{70}{3}-\frac{84}{3}=-\frac{14}{3}$$ Hence, $$\lambda+\mu=\frac{14}{3}-\frac{14}{3}=0$$ --- 5. **Sum of all possible values of $\lambda+\mu$** Possible values are: $$24,\ 0$$ Their sum is $$24+0=24$$ --- 6. **Evaluate options** - **A: 24** ✅ - **B: 0** ❌ - **C: 18** ❌ - **D: 6** ❌ Therefore, the correct option is **A**.

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