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The place $x-2 y+z=0$ is parallel to the line
Options:
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Verified Answer
The correct answer is:
$\frac{x-3}{4}=\frac{y-4}{5}=\frac{z-3}{6}$
Consider the equation of line given in option (a). The DR's of this line or $(4,5,6)$.
We know that if the line $\frac{x-x_0}{a_1}=\frac{y-y_0}{b_1}=\frac{z-z_0}{c_1}$ is parallel to the plane $a_2 x+b_2 y+c_2 z+d=0$, then $a_1 a_2+b_1 b_2+c_1 c_2=0$, that is the normal to the plane is perpendicular to the line.
Here, the vector $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is normal to the plane $x-2 y+z=0$
and $4(1)+5(-2)+6(1)=4-10+6=10-10=0$ So, option (a) is correct.
We know that if the line $\frac{x-x_0}{a_1}=\frac{y-y_0}{b_1}=\frac{z-z_0}{c_1}$ is parallel to the plane $a_2 x+b_2 y+c_2 z+d=0$, then $a_1 a_2+b_1 b_2+c_1 c_2=0$, that is the normal to the plane is perpendicular to the line.
Here, the vector $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is normal to the plane $x-2 y+z=0$
and $4(1)+5(-2)+6(1)=4-10+6=10-10=0$ So, option (a) is correct.
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