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Assertion (A) a, b, c, d are position vectors of 4 points such that $2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}-6 \mathbf{d}=0 \Rightarrow$ $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are coplanar.
Reason (R) Vector equation of the plane passing through three points whose position vectors are $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is $\mathbf{r}=(1-x-y) \mathbf{a}+x \mathbf{b}+y \mathbf{c}$.
Which of the following is true?
Options:
Reason (R) Vector equation of the plane passing through three points whose position vectors are $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is $\mathbf{r}=(1-x-y) \mathbf{a}+x \mathbf{b}+y \mathbf{c}$.
Which of the following is true?
Solution:
2168 Upvotes
Verified Answer
The correct answer is:
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of (A)
Vector equation of plane passing through three non-collinear points is
$$
\mathbf{r}=(1-x-y) \mathbf{a}+x \mathbf{b}+y \mathbf{c}
$$
If $a, b, c, d$ are coplanar, then $\mathbf{d}$ should satisfy
Eq. (i) for some $x, y$
$$
\mathbf{d}=(1-x-y) \mathbf{a}+x(\mathbf{b})+y \mathbf{c}
$$
Given, $\quad 2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}=6 \mathbf{d}$
$\Rightarrow \quad \frac{2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}}{6}=\mathbf{d}$
By substituting the value of $\mathbf{d}$ in Eq. (i), we get
$$
\frac{2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}}{6}=(1-x-y) \mathbf{a}+x(\mathbf{b})+y \mathbf{c}
$$
$$
\Rightarrow \quad 1-x-y=\frac{1}{3}
$$
$$
x=-\frac{1}{2}, y=\frac{7}{6}
$$
Substituting value of $x$ and $y$ in Eq. (ii)
$1-\left(-\frac{1}{2}\right)-\frac{7}{6}=\frac{1}{3}$ satisfy the Eq.(ii)
Hence, $\mathbf{b}$ lie on plane $\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are coplanar Hence, $A$ is correct, $R$ is correct and $R$ is correct explanation for $A$.
$$
\mathbf{r}=(1-x-y) \mathbf{a}+x \mathbf{b}+y \mathbf{c}
$$
If $a, b, c, d$ are coplanar, then $\mathbf{d}$ should satisfy
Eq. (i) for some $x, y$
$$
\mathbf{d}=(1-x-y) \mathbf{a}+x(\mathbf{b})+y \mathbf{c}
$$
Given, $\quad 2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}=6 \mathbf{d}$
$\Rightarrow \quad \frac{2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}}{6}=\mathbf{d}$
By substituting the value of $\mathbf{d}$ in Eq. (i), we get
$$
\frac{2 \mathbf{a}-3 \mathbf{b}+7 \mathbf{c}}{6}=(1-x-y) \mathbf{a}+x(\mathbf{b})+y \mathbf{c}
$$
$$
\Rightarrow \quad 1-x-y=\frac{1}{3}
$$
$$
x=-\frac{1}{2}, y=\frac{7}{6}
$$
Substituting value of $x$ and $y$ in Eq. (ii)
$1-\left(-\frac{1}{2}\right)-\frac{7}{6}=\frac{1}{3}$ satisfy the Eq.(ii)
Hence, $\mathbf{b}$ lie on plane $\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are coplanar Hence, $A$ is correct, $R$ is correct and $R$ is correct explanation for $A$.
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