If bar{r} cdot(2 bar{i}+3 bar{j}+4 bar{k})=5 | vedclass

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(B) The line of intersection of two planes $\bar{r} \cdot \bar{n}_1 = d_1$ and $\bar{r} \cdot \bar{n}_2 = d_2$ is parallel to the vector $\bar{v} = \b

Vedclass · Accepted Answer

$(16-7 \lambda) \bar{i}+(6 \lambda-9) \bar{j}-\lambda \bar{k}$

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(B) The line of intersection of two planes $\bar{r} \cdot \bar{n}_1 = d_1$ and $\bar{r} \cdot \bar{n}_2 = d_2$ is parallel to the vector $\bar{v} = \bar{n}_1 	imes \bar{n}_2$. Here, $\bar{n}_1 = 2 \bar{i} + 3 \bar{j} + 4 \bar{k}$ and $\bar{n}_2 = \bar{i} + \bar{j} - \bar{k}$. Calculating the cross product: $\bar{v} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 2 & 3 & 4 \ 1 & 1 & -1 \end{vmatrix} = \bar{i}(-3-4) - \bar{j}(-2-4) + \bar{k}(2-3) = -7 \bar{i} + 6 \bar{j} - \bar{k}$. The line passes through the point $(16, -9, 0)$, which corresponds to the position vector $\bar{a} = 16 \bar{i} - 9 \bar{j}$. The vector equation of a line is $\bar{r} = \bar{a} + \lambda \bar{v}$. Substituting the values: $\bar{r} = (16 \bar{i} - 9 \bar{j}) + \lambda(-7 \bar{i} + 6 \bar{j} - \bar{k}) = (16 - 7 \lambda) \bar{i} + (6 \lambda - 9) \bar{j} - \lambda \bar{k}$. Comparing this with the given options, option $B$ matches.

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\bar{r}=$

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