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  • Let the lines <img alt="\mathrm{L}_{1}: \overrightarrow{\mathbf{r}}=\lambda(\hat{i}+2 \hat{j}+3 \hat{k}), \lambda \in \mathbf{R}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cm

Let the lines

\mathrm{L}_{1}: \overrightarrow{\mathbf{r}}=\lambda(\hat{i}+2 \hat{j}+3 \hat{k}), \lambda \in \mathbf{R}

\mathrm{L}_{2}: \overrightarrow{\mathrm{r}}=(\hat{i}+3 \hat{j}+\hat{k})+\mu(\hat{i}+\hat{j}+5 \hat{k}) ; \mu \in \mathbf{R}

intersect at the point S. If a plane ax + by - z + d = 0 passes through S and is parallel to both the lines L_{1} and L_{2}, then the value of a+b+d is equal to____________.

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

L_{1}:\vec{r}=\lambda(\hat{i}+2 \hat{j}+3 \hat{k}), \lambda \in R

L_{2}: \vec{r}=(\hat{i}+3 \hat{j}+\hat{k})+\mu(\hat{i}+\hat{j}+5 \hat{k})

From the given conditions, the plane contains both the lines

Normal vector is cross product of direction vectors of both the lines

\vec{n} = (i + 2j + 3k) \times (i + j + 5k)
\vec{n} = (7i - 2j - k)

It also passes through point (0,0,0) lying on line 1

So the plane is 7(x - 0) -  2(y - 0) - (z - 0) = 0

Hence a = 7, b = -2, c = -1, d = 0

a + b + d = 5

 

Posted by

HARSH KANKARIA

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