Question

If for a > 0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is equal to ______.

Options

• `sqrt(41)`

• `sqrt(55)`

• `sqrt(31)`

• `sqrt(66)`

Answer 1

If for a > 0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is equal to `underlinebbsqrt(66)`.

Explanation:

CD = AM  ...(i)

In ΔABM: sin`phi = (AM)/|AB|`  ...(ii)

Using equation (i) and (ii)

AM = CD = |AB|sin`phi`

CD = `|AB|(sqrt(1 - cos^2phi))`

⇒ CD = `|AB| sqrt(1 - ((vec(AB).vecn)/|vec(AB)|)^2)`  ...`(∵ cosphi = (vec(AB).vecn)/(|vecn||vec(AB)|))`

⇒ CD = `sqrt((|vec(AB)|)^2 - (vec(AB).vecn)^2)`  ...(iii)

As A(a, –2a, 3) and B(0, 4, 5) then `vec(AB) = vec(OB) - vec(OA) = - ahati + (2a + 4)hatj + 2hatk`

`vec(AB).vecn` = –la + (2a + 4)m + 2n  ...(iv)

Since, C(0, –a, –1) lies on plane lx + my + nz = 0,

So, 0l – am – n = 0 ⇒ `m/n = (-1)/a`

From the figure

`vec(AC)||vecn`

`a/l = (-a)/m = 4/n`

m = –l and `m/n = (-a)/4`  ...(vi)

Now using equations (v) and (vi)

⇒ a² = 4

a = ±2

As a > 0, a = 2

Now from equation (vii)

2m + n = 0  [As l² + m² + n² = 1] ...(vii)

∴ m² + m² + 4m² = 1

∴ m² = `1/6`

∴ m = `+-1/6`

∴ m = `1/sqrt(6)`

Using equation (vii)

n = –2m

n = `(-2)/sqrt(6)`

l = `(-1)/sqrt(6)`

Now from equation (iv)

`vec(AB).vecn = -2((-1)/sqrt(6)) + 8(1/sqrt(6)) + 2((-2)/sqrt(6))`

= `(+2 + 8 - 4)/sqrt(6)`

= `sqrt(6)`

`vec(AB) = sqrt(a^2 + (2a + 4)^2 + (2)^2`

= `sqrt(2^2 + 8^2 + 4^2)`

`|vec(AB)| = sqrt(4 + 64 + 4) = sqrt(72)`

CD = `sqrt((sqrt(72))^2 - (sqrt(6))^2`

CD = `sqrt(72 - 6)`

CD = `sqrt(66)`