Let \overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k and \overrightarrow c = 2\widehat i + 4\widehat j + ( \lambda ^2 - 1 )\widehat k be coplanar vectors. Then the non-zero vector \overrightarrow a \times \overrightarrow c is :

Vector Algebra PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q30 with Solution

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Correct answer: B

Write the vectors in component form

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

Since the three vectors are coplanar, their scalar triple product must be zero:

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

That is,

1 & 2 & 4\\ 1 & \lambda & 4\\ 2 & 4 & \lambda^2-1 \end{vmatrix}=0$$ 2. **Evaluate the determinant** Expanding along the first row:

\begin{aligned} \Delta &= 1\begin{vmatrix} \lambda & 4 \ 4 & \lambda^2-1 \end{vmatrix} -2\begin{vmatrix} 1 & 4 \ 2 & \lambda^2-1 \end{vmatrix} +4\begin{vmatrix} 1 & \lambda \ 2 & 4 \end{vmatrix} \[4pt] &=1\big(\lambda(\lambda^2-1)-16\big)-2\big((\lambda^2-1)-8\big)+4(4-2\lambda) \[4pt] &=\lambda^3-\lambda-16-2\lambda^2+18+16-8\lambda \[4pt] &=\lambda^3-2\lambda^2-9\lambda+18 \end{aligned}

So, $$\lambda^3-2\lambda^2-9\lambda+18=0$$ 3. **Factorize** Try rational roots. For $\lambda=2$: $$8-8-18+18=0$$ So $(\lambda-2)$ is a factor. Divide: $$\lambda^3-2\lambda^2-9\lambda+18=(\lambda-2)(\lambda^2-9)=(\lambda-2)(\lambda-3)(\lambda+3)$$ Hence, $$\lambda=2,3,-3$$ But note that if $\lambda=2$, then $$\vec b=(1,2,4)=\vec a$$ The question asks for the **non-zero** vector $\vec a\times \vec c$, so let us compute it and check which value gives a non-zero result. 4. **Compute $\vec a\times \vec c$**

\vec a\times \vec c= \begin{vmatrix} \hat i & \hat j & \hat k\ 1 & 2 & 4\ 2 & 4 & \lambda^2-1 \end{vmatrix}

\begin{aligned} \vec a\times \vec c &=\hat i\big(2(\lambda^2-1)-16\big)-\hat j\big((\lambda^2-1)-8\big)+\hat k(4-4) \[4pt] &=(2\lambda^2-18)\hat i-(\lambda^2-9)\hat j \[4pt] &=2(\lambda^2-9)\hat i-(\lambda^2-9)\hat j \[4pt] &=(\lambda^2-9)(2\hat i-\hat j) \end{aligned}

5. **Use the possible values of $\lambda$** - If $\lambda=2$, then $\lambda^2-9=4-9=-5$: $$\vec a\times \vec c=-5(2\hat i-\hat j)=-10\hat i+5\hat j$$ - If $\lambda=3$ or $\lambda=-3$, then $\lambda^2-9=0$, so $$\vec a\times \vec c=\vec 0$$ Since the question specifies **non-zero vector** $\vec a\times \vec c$, only $\lambda=2$ is valid. Therefore, $$\boxed{\vec a\times \vec c=-10\hat i+5\hat j}$$ 6. **Match with the options** This is **Option B**.

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View written solutionFree

Correct answer: B

Write the vectors in component form

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

Since the three vectors are coplanar, their scalar triple product must be zero:

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

That is,

1 & 2 & 4\\ 1 & \lambda & 4\\ 2 & 4 & \lambda^2-1 \end{vmatrix}=0$$ 2. **Evaluate the determinant** Expanding along the first row:

So, $$\lambda^3-2\lambda^2-9\lambda+18=0$$ 3. **Factorize** Try rational roots. For $\lambda=2$: $$8-8-18+18=0$$ So $(\lambda-2)$ is a factor. Divide: $$\lambda^3-2\lambda^2-9\lambda+18=(\lambda-2)(\lambda^2-9)=(\lambda-2)(\lambda-3)(\lambda+3)$$ Hence, $$\lambda=2,3,-3$$ But note that if $\lambda=2$, then $$\vec b=(1,2,4)=\vec a$$ The question asks for the **non-zero** vector $\vec a\times \vec c$, so let us compute it and check which value gives a non-zero result. 4. **Compute $\vec a\times \vec c$**

\vec a\times \vec c= \begin{vmatrix} \hat i & \hat j & \hat k\ 1 & 2 & 4\ 2 & 4 & \lambda^2-1 \end{vmatrix}

5. **Use the possible values of $\lambda$** - If $\lambda=2$, then $\lambda^2-9=4-9=-5$: $$\vec a\times \vec c=-5(2\hat i-\hat j)=-10\hat i+5\hat j$$ - If $\lambda=3$ or $\lambda=-3$, then $\lambda^2-9=0$, so $$\vec a\times \vec c=\vec 0$$ Since the question specifies **non-zero vector** $\vec a\times \vec c$, only $\lambda=2$ is valid. Therefore, $$\boxed{\vec a\times \vec c=-10\hat i+5\hat j}$$ 6. **Match with the options** This is **Option B**.

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