Let a straight line L pass through the point P(2,−1,3) and be perpendicular to the lines 4x−1=4y+1=2z−3 and 4x−3=3y−2=4z+2. If the line L intersects the yz -plane at the point Q, then the distance between the points P and Q is:
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Step-by-Step Solution
Step 1: Find the direction vector of line L
The line L is perpendicular to two given lines. The direction vectors of the given lines are d1=⟨4,4,2⟩ and d2=⟨4,3,4⟩. The direction vector of line L, denoted as dL, must be perpendicular to both d1 and d2. Therefore, dL can be found by taking the cross product of d1 and d2.
Step 2: Calculate the cross product
We calculate the determinant to find the components of the cross product. This gives us the direction vector dL=⟨10,−8,−4⟩. We can simplify this direction vector by dividing by a common factor of 2, so dL=⟨5,−4,−2⟩.
Step 3: Write the equation of line L
The line L passes through the point P(2,−1,3) and has the direction vector dL=⟨5,−4,−2⟩. We can write the symmetric form of the equation of line L using these values. We introduce a parameter λ to represent any point on the line.
Step 4: Find the point Q where L intersects the yz-plane
The yz -plane is defined by x=0. We substitute x=0 into the parametric equations of line L to find the value of λ. Once λ is found, we substitute it back into the parametric equations to get the coordinates of point Q.
Step 5: Calculate the distance between P and Q
Finally, we use the distance formula to find the distance between point P(2,−1,3) and point Q(0,53,519). We substitute the coordinates into the formula and simplify the expression to get the final distance.