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Good evening Janet! Here are my workings for part b.
To do this question, we have to define a point F on the line such that PF is perpendicular to the line (something like the O Level’s shortest distance from a point to a line).
Connecting P and F together, we have PF being the direction vector of the line (similar to the “gradient” idea in O Level).
Line L is defined by a point and a direction vector (similar to equation of line given a point and a gradient in O Level). For now, this direction vector (“gradient”) is of importance.
Because PF is perpendicular to line L, the dot product of their direction vectors must equal to zero (slightly different from O Levels, in which products of gradients is -1).
From here, we proceed to obtain the specific value of lambda (that non-English symbol).
To do this question, we have to define a point F on the line such that PF is perpendicular to the line (something like the O Level’s shortest distance from a point to a line).
Connecting P and F together, we have PF being the direction vector of the line (similar to the “gradient” idea in O Level).
Line L is defined by a point and a direction vector (similar to equation of line given a point and a gradient in O Level). For now, this direction vector (“gradient”) is of importance.
Because PF is perpendicular to line L, the dot product of their direction vectors must equal to zero (slightly different from O Levels, in which products of gradients is -1).
From here, we proceed to obtain the specific value of lambda (that non-English symbol).
Date Posted:
4 years ago
Thank you
done
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Good evening Janet! Here are my workings for part c.
Date Posted:
4 years ago